SQL query to find a list of city names that dont start with vowels

I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population

This is the query that I've written

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

This gives a wrong answer. What am I doing wrong here?

asked Apr 14 '16 at 21:59

Zeus

7832829

try the NOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05

add a comment |

I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population

This is the query that I've written

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

This gives a wrong answer. What am I doing wrong here?

asked Apr 14 '16 at 21:59

Zeus

7832829

try the NOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05

add a comment |

I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population

This is the query that I've written

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

This gives a wrong answer. What am I doing wrong here?

asked Apr 14 '16 at 21:59

Zeus

7832829

I'm trying to query the list of CITY names from table - STATION that do not start with vowels with results that cannot contain duplicates.
The table just has id, city, population

This is the query that I've written

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '[^bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

This gives a wrong answer. What am I doing wrong here?

mysql

asked Apr 14 '16 at 21:59

Zeus

7832829

asked Apr 14 '16 at 21:59

Zeus

7832829

asked Apr 14 '16 at 21:59

Zeus

7832829

asked Apr 14 '16 at 21:59

Zeus

7832829

asked Apr 14 '16 at 21:59

Zeus

7832829

try the NOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05

add a comment |

try the NOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05

try the NOT RLIKE '^[aeiouAEIOU].*$'
– Tin Tran
Apr 14 '16 at 22:05

add a comment |

25 Answers
25

active

oldest

votes

try this.

SELECT DISTINCT CITY 

FROM STATION 

WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

add a comment |

A ^ in regular expressions can have different meanings, depending on its location. When it's the first character in a regex, it refers to the start of the string. But when it's the first character in a set, like [^abc], it means not one of. And when it appears elsewhere, it just refers to the ^ itself.

So you would need something like:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

or, just exclude the letters you don't want:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[^aeiouAEIOU].*';

answered Apr 14 '16 at 22:07

Arjan

8,34812032

add a comment |

Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE

select distinct city 

from station 

where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');

In MySQL -

 select distinct city 

 from station 

 where city RLIKE '^[^aeiouAEIOU].*' OR 

 city RLIKE '^.*[^aeiouAEIOU]$';

edited Aug 6 '18 at 18:00

Community♦

answered Feb 18 '18 at 18:45

Rahul Yadav

233

add a comment |

select distinct city from station 

where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and

      city Not like 'o%' and city not like 'U%';

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

answered Jul 14 '16 at 9:58

satish

211

add a comment |

This worked on MS SQL SERVER:

select distinct city from station where city NOT like '[aeuio]%' Order by City;

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

answered Dec 24 '16 at 8:41

abhinay vijay

311

it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41

add a comment |

SELECT DISTINCT CITY from STATION  WHERE CITY RLIKE '^[^aeiouAEIOU].*$'

edited Dec 24 '16 at 8:59

Pang

6,8911563101

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48

add a comment |

select distinct city from station where city regexp

'^[^aeiou].*[^aeiou]$'

If your query doesn't starts and ends with a vowel.

answered May 21 '17 at 16:42

marwari

587

add a comment |

Try the following:

SELECT DISTINCT CITY FROM STATION

WHERE CITY REGEXP '^[^aeiou].*';

MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

answered Feb 21 '18 at 16:42

simran kumari

111

add a comment |

Simple example without regular expressions

SELECT DISTINCT 

    City 

FROM Station

WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

answered Sep 3 '18 at 11:04

ed lovejoy

111

add a comment |

I tried this.

select distinct city

from station

where substring(city,1,1) not in('A','E','I','O','U');

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

answered Dec 5 '18 at 20:19

Makhmud Galy

112

add a comment |

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

add a comment |

In MS SQL:

Select DISTINCT CITY

FROM

STATION

Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';

answered Jan 24 '17 at 23:59

Krishneil

630816

add a comment |

you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'

answered Mar 8 '17 at 4:33

Sajal Singh

146211

add a comment |

select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');

answered Mar 8 '17 at 17:42

Mohit

571212

add a comment |

select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'

answered May 26 '17 at 10:14

Aman Verma

614

add a comment |

try this one

SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

add a comment |

Select Distinct City from Station where City Like '[^aeiou]%';

MS SQL Server Tested.

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

answered Apr 5 '18 at 21:02

Darpan

add a comment |

 SELECT DISTINCT city from STATION

 WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"

 AND city NOT LIKE "[aieuo]%[aieuo]";

The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.

answered Sep 25 '18 at 20:05

David Osipov

add a comment |

select distinct city from station where (

    left(city,1) not in ('a','e','i','o','u') 

    and  

    right(city,1) not in ('a','e','i','o','u')

);

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

answered Oct 21 '18 at 9:27

Rohit Rawat

1

PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05

add a comment |

In ORACLE:

select 

distinct(city)

from station

where regexp_like (city, '^[^A|E|I|O|U](*)');

There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

answered Nov 19 '18 at 20:40

Aaron W.

Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41

add a comment |

for oracle

select distinct 

    city 

from station 

where regexp_like(city, '^[^aeiouAEIOU].*');

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

answered Apr 28 '18 at 19:47

Rasim SEN

add a comment |

select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'

distinct: to avoid the duplication on the following column

regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.

[^...] means any character NOT contain any of ...
[^...]. means a single character NOT contain any of ...
[^...].* means first character NOT contain any of ...
[^...].*[^...] means first character NOT contains any of ... AND not ends with ...
[^aeiou].*[^aeiou] mean the first character NOT starts with vowels AND not ends with vowels

edited Nov 26 '18 at 3:09

answered Nov 25 '18 at 3:11

Jay

31226

While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28

add a comment |

Mysql we can use this.

select distinct city from station where city regexp '^[^aeiou].*';

To Know More About MySQL Regular Expression

For Oracle we can use

select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');

To Know More About Oracle Regular Expression

For MS SQL Server

select distinct city from station where city not like '[aeuio]%';

answered Nov 30 '18 at 4:22

Arun Solomon

112

add a comment |

-1

select distinct city from station where

city not in

(select distinct city from station where

 (city like "A%" or

city like "E%" or

 city like "I%" or

 city like "O%" or

 city like "U%"

));

This will work i guess.

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

answered Oct 18 '16 at 3:04

Saikrishna Syamala

please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27

add a comment |

-1

SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'

This works for MYSQL

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11

Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f36634742%2fsql-query-to-find-a-list-of-city-names-that-dont-start-with-vowels%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

25 Answers
25

active

oldest

votes

25 Answers
25

active

oldest

votes

try this.

SELECT DISTINCT CITY 

FROM STATION 

WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

add a comment |

try this.

SELECT DISTINCT CITY 

FROM STATION 

WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

add a comment |

try this.

SELECT DISTINCT CITY 

FROM STATION 

WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

try this.

SELECT DISTINCT CITY 

FROM STATION 

WHERE CITY NOT RLIKE '^[aeiouAEIOU].*$'

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

answered Apr 14 '16 at 22:06

Tin Tran

5,41421226

add a comment |

So you would need something like:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

or, just exclude the letters you don't want:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[^aeiouAEIOU].*';

answered Apr 14 '16 at 22:07

Arjan

8,34812032

add a comment |

So you would need something like:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

or, just exclude the letters you don't want:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[^aeiouAEIOU].*';

answered Apr 14 '16 at 22:07

Arjan

8,34812032

add a comment |

So you would need something like:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

or, just exclude the letters you don't want:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[^aeiouAEIOU].*';

answered Apr 14 '16 at 22:07

Arjan

8,34812032

So you would need something like:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ].*';

or, just exclude the letters you don't want:

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^[^aeiouAEIOU].*';

answered Apr 14 '16 at 22:07

Arjan

8,34812032

answered Apr 14 '16 at 22:07

Arjan

8,34812032

answered Apr 14 '16 at 22:07

Arjan

8,34812032

answered Apr 14 '16 at 22:07

Arjan

8,34812032

add a comment |

Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE

select distinct city 

from station 

where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');

In MySQL -

 select distinct city 

 from station 

 where city RLIKE '^[^aeiouAEIOU].*' OR 

 city RLIKE '^.*[^aeiouAEIOU]$';

edited Aug 6 '18 at 18:00

Community♦

answered Feb 18 '18 at 18:45

Rahul Yadav

233

add a comment |

Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE

select distinct city 

from station 

where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');

In MySQL -

 select distinct city 

 from station 

 where city RLIKE '^[^aeiouAEIOU].*' OR 

 city RLIKE '^.*[^aeiouAEIOU]$';

edited Aug 6 '18 at 18:00

Community♦

answered Feb 18 '18 at 18:45

Rahul Yadav

233

add a comment |

Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE

select distinct city 

from station 

where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');

In MySQL -

 select distinct city 

 from station 

 where city RLIKE '^[^aeiouAEIOU].*' OR 

 city RLIKE '^.*[^aeiouAEIOU]$';

edited Aug 6 '18 at 18:00

Community♦

answered Feb 18 '18 at 18:45

Rahul Yadav

233

Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. IN ORACLE

select distinct city 

from station 

where regexp_like(city, '^[^aeiouAEIOU]|*[^aeiouAEIOU]$');

In MySQL -

 select distinct city 

 from station 

 where city RLIKE '^[^aeiouAEIOU].*' OR 

 city RLIKE '^.*[^aeiouAEIOU]$';

edited Aug 6 '18 at 18:00

Community♦

answered Feb 18 '18 at 18:45

Rahul Yadav

233

edited Aug 6 '18 at 18:00

Community♦

edited Aug 6 '18 at 18:00

Community♦

edited Aug 6 '18 at 18:00

Community♦

answered Feb 18 '18 at 18:45

Rahul Yadav

233

answered Feb 18 '18 at 18:45

Rahul Yadav

233

answered Feb 18 '18 at 18:45

Rahul Yadav

233

add a comment |

select distinct city from station 

where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and

      city Not like 'o%' and city not like 'U%';

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

answered Jul 14 '16 at 9:58

satish

211

add a comment |

select distinct city from station 

where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and

      city Not like 'o%' and city not like 'U%';

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

answered Jul 14 '16 at 9:58

satish

211

add a comment |

select distinct city from station 

where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and

      city Not like 'o%' and city not like 'U%';

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

answered Jul 14 '16 at 9:58

satish

211

select distinct city from station 

where city Not like 'A%' and city Not like 'E%' and city Not like 'I%' and

      city Not like 'o%' and city not like 'U%';

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

answered Jul 14 '16 at 9:58

satish

211

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

edited Jul 14 '16 at 12:41

Bo Persson

77.6k17116184

answered Jul 14 '16 at 9:58

satish

211

answered Jul 14 '16 at 9:58

satish

211

answered Jul 14 '16 at 9:58

satish

211

add a comment |

This worked on MS SQL SERVER:

select distinct city from station where city NOT like '[aeuio]%' Order by City;

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

answered Dec 24 '16 at 8:41

abhinay vijay

311

it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41

add a comment |

This worked on MS SQL SERVER:

select distinct city from station where city NOT like '[aeuio]%' Order by City;

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

answered Dec 24 '16 at 8:41

abhinay vijay

311

it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41

add a comment |

This worked on MS SQL SERVER:

select distinct city from station where city NOT like '[aeuio]%' Order by City;

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

answered Dec 24 '16 at 8:41

abhinay vijay

311

This worked on MS SQL SERVER:

select distinct city from station where city NOT like '[aeuio]%' Order by City;

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

answered Dec 24 '16 at 8:41

abhinay vijay

311

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

edited Dec 24 '16 at 8:56

FelixSFD

4,27572986

answered Dec 24 '16 at 8:41

abhinay vijay

311

answered Dec 24 '16 at 8:41

abhinay vijay

311

answered Dec 24 '16 at 8:41

abhinay vijay

311

it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41

add a comment |

it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41

it worked on MS SQL SERVER
– abhinay vijay
Dec 24 '16 at 8:41

add a comment |

SELECT DISTINCT CITY from STATION  WHERE CITY RLIKE '^[^aeiouAEIOU].*$'

edited Dec 24 '16 at 8:59

Pang

6,8911563101

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48

add a comment |

SELECT DISTINCT CITY from STATION  WHERE CITY RLIKE '^[^aeiouAEIOU].*$'

edited Dec 24 '16 at 8:59

Pang

6,8911563101

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48

add a comment |

SELECT DISTINCT CITY from STATION  WHERE CITY RLIKE '^[^aeiouAEIOU].*$'

edited Dec 24 '16 at 8:59

Pang

6,8911563101

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

SELECT DISTINCT CITY from STATION  WHERE CITY RLIKE '^[^aeiouAEIOU].*$'

edited Dec 24 '16 at 8:59

Pang

6,8911563101

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

edited Dec 24 '16 at 8:59

Pang

6,8911563101

edited Dec 24 '16 at 8:59

Pang

6,8911563101

edited Dec 24 '16 at 8:59

Pang

6,8911563101

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

answered Sep 16 '16 at 0:47

SunnyShineColorado

111

This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48

add a comment |

This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48

This works for mysql
– SunnyShineColorado
Sep 16 '16 at 0:48

add a comment |

select distinct city from station where city regexp

'^[^aeiou].*[^aeiou]$'

If your query doesn't starts and ends with a vowel.

answered May 21 '17 at 16:42

marwari

587

add a comment |

select distinct city from station where city regexp

'^[^aeiou].*[^aeiou]$'

If your query doesn't starts and ends with a vowel.

answered May 21 '17 at 16:42

marwari

587

add a comment |

select distinct city from station where city regexp

'^[^aeiou].*[^aeiou]$'

If your query doesn't starts and ends with a vowel.

answered May 21 '17 at 16:42

marwari

587

select distinct city from station where city regexp

'^[^aeiou].*[^aeiou]$'

If your query doesn't starts and ends with a vowel.

answered May 21 '17 at 16:42

marwari

587

answered May 21 '17 at 16:42

marwari

587

answered May 21 '17 at 16:42

marwari

587

answered May 21 '17 at 16:42

marwari

587

add a comment |

Try the following:

SELECT DISTINCT CITY FROM STATION

WHERE CITY REGEXP '^[^aeiou].*';

MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

answered Feb 21 '18 at 16:42

simran kumari

111

add a comment |

Try the following:

SELECT DISTINCT CITY FROM STATION

WHERE CITY REGEXP '^[^aeiou].*';

MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

answered Feb 21 '18 at 16:42

simran kumari

111

add a comment |

Try the following:

SELECT DISTINCT CITY FROM STATION

WHERE CITY REGEXP '^[^aeiou].*';

MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

answered Feb 21 '18 at 16:42

simran kumari

111

Try the following:

SELECT DISTINCT CITY FROM STATION

WHERE CITY REGEXP '^[^aeiou].*';

MySQL Ref.: https://dev.mysql.com/doc/refman/5.5/en/regexp.html

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

answered Feb 21 '18 at 16:42

simran kumari

111

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

edited Feb 21 '18 at 17:59

Syed Aslam

7,62953153

answered Feb 21 '18 at 16:42

simran kumari

111

answered Feb 21 '18 at 16:42

simran kumari

111

answered Feb 21 '18 at 16:42

simran kumari

111

add a comment |

Simple example without regular expressions

SELECT DISTINCT 

    City 

FROM Station

WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

answered Sep 3 '18 at 11:04

ed lovejoy

111

add a comment |

Simple example without regular expressions

SELECT DISTINCT 

    City 

FROM Station

WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

answered Sep 3 '18 at 11:04

ed lovejoy

111

add a comment |

Simple example without regular expressions

SELECT DISTINCT 

    City 

FROM Station

WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

answered Sep 3 '18 at 11:04

ed lovejoy

111

Simple example without regular expressions

SELECT DISTINCT 

    City 

FROM Station

WHERE LEFT(City, 1) NOT IN ("a", "e", "i", "o", "u");

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

answered Sep 3 '18 at 11:04

ed lovejoy

111

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

edited Nov 25 '18 at 3:55

digital.aaron

3,1341128

answered Sep 3 '18 at 11:04

ed lovejoy

111

answered Sep 3 '18 at 11:04

ed lovejoy

111

answered Sep 3 '18 at 11:04

ed lovejoy

111

add a comment |

I tried this.

select distinct city

from station

where substring(city,1,1) not in('A','E','I','O','U');

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

answered Dec 5 '18 at 20:19

Makhmud Galy

112

add a comment |

I tried this.

select distinct city

from station

where substring(city,1,1) not in('A','E','I','O','U');

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

answered Dec 5 '18 at 20:19

Makhmud Galy

112

add a comment |

I tried this.

select distinct city

from station

where substring(city,1,1) not in('A','E','I','O','U');

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

answered Dec 5 '18 at 20:19

Makhmud Galy

112

I tried this.

select distinct city

from station

where substring(city,1,1) not in('A','E','I','O','U');

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

answered Dec 5 '18 at 20:19

Makhmud Galy

112

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

edited Dec 5 '18 at 20:42

John Montgomery

2,44961937

answered Dec 5 '18 at 20:19

Makhmud Galy

112

answered Dec 5 '18 at 20:19

Makhmud Galy

112

answered Dec 5 '18 at 20:19

Makhmud Galy

112

add a comment |

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

add a comment |

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

add a comment |

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

SELECT DISTINCT CITY FROM STATION 

WHERE CITY RLIKE '^.*[bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]$';

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

edited Dec 9 '16 at 17:25

Paul Roub

32.6k85773

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

answered Dec 9 '16 at 17:16

MANJUNATH KAMMAR

add a comment |

In MS SQL:

Select DISTINCT CITY

FROM

STATION

Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';

answered Jan 24 '17 at 23:59

Krishneil

630816

add a comment |

In MS SQL:

Select DISTINCT CITY

FROM

STATION

Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';

answered Jan 24 '17 at 23:59

Krishneil

630816

add a comment |

In MS SQL:

Select DISTINCT CITY

FROM

STATION

Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';

answered Jan 24 '17 at 23:59

Krishneil

630816

In MS SQL:

Select DISTINCT CITY

FROM

STATION

Where CITY NOT LIKE 'A%' AND CITY NOT LIKE 'E%' AND CITY NOT LIKE 'I%' AND CITY NOT LIKE 'O%' AND CITY NOT LIKE 'U%';

answered Jan 24 '17 at 23:59

Krishneil

630816

answered Jan 24 '17 at 23:59

Krishneil

630816

answered Jan 24 '17 at 23:59

Krishneil

630816

answered Jan 24 '17 at 23:59

Krishneil

630816

add a comment |

you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'

answered Mar 8 '17 at 4:33

Sajal Singh

146211

add a comment |

you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'

answered Mar 8 '17 at 4:33

Sajal Singh

146211

add a comment |

you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'

answered Mar 8 '17 at 4:33

Sajal Singh

146211

you can use
select distinct city from station where city not RLIKE '^[aeiouAEIOU].*'

answered Mar 8 '17 at 4:33

Sajal Singh

146211

answered Mar 8 '17 at 4:33

Sajal Singh

146211

answered Mar 8 '17 at 4:33

Sajal Singh

146211

answered Mar 8 '17 at 4:33

Sajal Singh

146211

add a comment |

select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');

answered Mar 8 '17 at 17:42

Mohit

571212

add a comment |

select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');

answered Mar 8 '17 at 17:42

Mohit

571212

add a comment |

select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');

answered Mar 8 '17 at 17:42

Mohit

571212

select distinct CITY from STATION where left(CITY,1) not in ('a','e','i','o','u') or right(CITY,1) not in ('a','e','i','o','u');

answered Mar 8 '17 at 17:42

Mohit

571212

answered Mar 8 '17 at 17:42

Mohit

571212

answered Mar 8 '17 at 17:42

Mohit

571212

answered Mar 8 '17 at 17:42

Mohit

571212

add a comment |

select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'

answered May 26 '17 at 10:14

Aman Verma

614

add a comment |

select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'

answered May 26 '17 at 10:14

Aman Verma

614

add a comment |

select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'

answered May 26 '17 at 10:14

Aman Verma

614

select DISTINCT CITY from STATION where CITY NOT LIKE '[a,e,i,o,u]%'

answered May 26 '17 at 10:14

Aman Verma

614

answered May 26 '17 at 10:14

Aman Verma

614

answered May 26 '17 at 10:14

Aman Verma

614

answered May 26 '17 at 10:14

Aman Verma

614

add a comment |

try this one

SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

add a comment |

try this one

SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

add a comment |

try this one

SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

try this one

SELECT DISTINCT CITY FROM STATION WHERE
LOWER(SUBSTR(CITY,LENGTH(CITY),1)) IN ('a','e','i','o','u');

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

answered Oct 12 '17 at 11:12

Ganesh Giri

1855

add a comment |

Select Distinct City from Station where City Like '[^aeiou]%';

MS SQL Server Tested.

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

answered Apr 5 '18 at 21:02

Darpan

add a comment |

Select Distinct City from Station where City Like '[^aeiou]%';

MS SQL Server Tested.

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

answered Apr 5 '18 at 21:02

Darpan

add a comment |

Select Distinct City from Station where City Like '[^aeiou]%';

MS SQL Server Tested.

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

answered Apr 5 '18 at 21:02

Darpan

Select Distinct City from Station where City Like '[^aeiou]%';

MS SQL Server Tested.

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

answered Apr 5 '18 at 21:02

Darpan

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

edited Apr 5 '18 at 21:32

Jose Da Silva

2,08421026

answered Apr 5 '18 at 21:02

Darpan

answered Apr 5 '18 at 21:02

Darpan

answered Apr 5 '18 at 21:02

Darpan

add a comment |

 SELECT DISTINCT city from STATION

 WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"

 AND city NOT LIKE "[aieuo]%[aieuo]";

The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.

answered Sep 25 '18 at 20:05

David Osipov

add a comment |

 SELECT DISTINCT city from STATION

 WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"

 AND city NOT LIKE "[aieuo]%[aieuo]";

The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.

answered Sep 25 '18 at 20:05

David Osipov

add a comment |

 SELECT DISTINCT city from STATION

 WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"

 AND city NOT LIKE "[aieuo]%[aieuo]";

The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.

answered Sep 25 '18 at 20:05

David Osipov

 SELECT DISTINCT city from STATION

 WHERE city NOT LIKE "[aieuo]%[aieuo] [aieuo]%[aieuo]"

 AND city NOT LIKE "[aieuo]%[aieuo]";

The first NOT LIKE means, that each word of the two-word city names doesn't start and end with vowels. The second NOT LIKE is for one-word city names.

answered Sep 25 '18 at 20:05

David Osipov

answered Sep 25 '18 at 20:05

David Osipov

answered Sep 25 '18 at 20:05

David Osipov

answered Sep 25 '18 at 20:05

David Osipov

add a comment |

select distinct city from station where (

    left(city,1) not in ('a','e','i','o','u') 

    and  

    right(city,1) not in ('a','e','i','o','u')

);

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

answered Oct 21 '18 at 9:27

Rohit Rawat

1

PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05

add a comment |

select distinct city from station where (

    left(city,1) not in ('a','e','i','o','u') 

    and  

    right(city,1) not in ('a','e','i','o','u')

);

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

answered Oct 21 '18 at 9:27

Rohit Rawat

1

PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05

add a comment |

select distinct city from station where (

    left(city,1) not in ('a','e','i','o','u') 

    and  

    right(city,1) not in ('a','e','i','o','u')

);

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

answered Oct 21 '18 at 9:27

Rohit Rawat

select distinct city from station where (

    left(city,1) not in ('a','e','i','o','u') 

    and  

    right(city,1) not in ('a','e','i','o','u')

);

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

answered Oct 21 '18 at 9:27

Rohit Rawat

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

edited Oct 21 '18 at 9:49

Mohammad

15.4k123361

answered Oct 21 '18 at 9:27

Rohit Rawat

answered Oct 21 '18 at 9:27

Rohit Rawat

answered Oct 21 '18 at 9:27

Rohit Rawat

1

PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05

add a comment |

1

PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05

PLease add some explanation why this code answers the question.
– Hintham
Oct 21 '18 at 9:34

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
– FrankerZ
Oct 21 '18 at 16:05

add a comment |

In ORACLE:

select 

distinct(city)

from station

where regexp_like (city, '^[^A|E|I|O|U](*)');

There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

answered Nov 19 '18 at 20:40

Aaron W.

Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41

add a comment |

In ORACLE:

select 

distinct(city)

from station

where regexp_like (city, '^[^A|E|I|O|U](*)');

There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

answered Nov 19 '18 at 20:40

Aaron W.

Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41

add a comment |

In ORACLE:

select 

distinct(city)

from station

where regexp_like (city, '^[^A|E|I|O|U](*)');

There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

answered Nov 19 '18 at 20:40

Aaron W.

In ORACLE:

select 

distinct(city)

from station

where regexp_like (city, '^[^A|E|I|O|U](*)');

There is no need to specify the ending letter in this solution. It's not called out for in the problem presented.

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

answered Nov 19 '18 at 20:40

Aaron W.

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

edited Nov 19 '18 at 20:50

Hossein Golshani

2,05851023

answered Nov 19 '18 at 20:40

Aaron W.

answered Nov 19 '18 at 20:40

Aaron W.

answered Nov 19 '18 at 20:40

Aaron W.

Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41

add a comment |

Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41

Between the two carots ( ^^ ) and after the U, there should be [ ]. They didn't post for some reason.
– Aaron W.
Nov 19 '18 at 20:41

add a comment |

for oracle

select distinct 

    city 

from station 

where regexp_like(city, '^[^aeiouAEIOU].*');

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

answered Apr 28 '18 at 19:47

Rasim SEN

add a comment |

for oracle

select distinct 

    city 

from station 

where regexp_like(city, '^[^aeiouAEIOU].*');

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

answered Apr 28 '18 at 19:47

Rasim SEN

add a comment |

for oracle

select distinct 

    city 

from station 

where regexp_like(city, '^[^aeiouAEIOU].*');

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

answered Apr 28 '18 at 19:47

Rasim SEN

for oracle

select distinct 

    city 

from station 

where regexp_like(city, '^[^aeiouAEIOU].*');

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

answered Apr 28 '18 at 19:47

Rasim SEN

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

edited Nov 25 '18 at 3:56

digital.aaron

3,1341128

answered Apr 28 '18 at 19:47

Rasim SEN

answered Apr 28 '18 at 19:47

Rasim SEN

answered Apr 28 '18 at 19:47

Rasim SEN

add a comment |

select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'

distinct: to avoid the duplication on the following column

regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.

edited Nov 26 '18 at 3:09

answered Nov 25 '18 at 3:11

Jay

31226

While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28

add a comment |

select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'

distinct: to avoid the duplication on the following column

regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.

edited Nov 26 '18 at 3:09

answered Nov 25 '18 at 3:11

Jay

31226

While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28

add a comment |

select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'

distinct: to avoid the duplication on the following column

regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.

edited Nov 26 '18 at 3:09

answered Nov 25 '18 at 3:11

Jay

31226

select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'

distinct: to avoid the duplication on the following column

regexp: MySql function (regexp) to obtain Regular Expressions. expression starts with ^ and ends with $. In Oracle its regexp_like (city,'RegEx') Regular Expressions With Oracle Database.

edited Nov 26 '18 at 3:09

answered Nov 25 '18 at 3:11

Jay

31226

edited Nov 26 '18 at 3:09

answered Nov 25 '18 at 3:11

Jay

31226

answered Nov 25 '18 at 3:11

Jay

31226

answered Nov 25 '18 at 3:11

Jay

31226

While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28

add a comment |

While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28

While this code may answer the question, providing additional context regarding how and why it solves the problem would improve the answer's long-term value.
– Alexander
Nov 25 '18 at 11:28

add a comment |

Mysql we can use this.

select distinct city from station where city regexp '^[^aeiou].*';

To Know More About MySQL Regular Expression

For Oracle we can use

select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');

To Know More About Oracle Regular Expression

For MS SQL Server

select distinct city from station where city not like '[aeuio]%';

answered Nov 30 '18 at 4:22

Arun Solomon

112

add a comment |

Mysql we can use this.

select distinct city from station where city regexp '^[^aeiou].*';

To Know More About MySQL Regular Expression

For Oracle we can use

select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');

To Know More About Oracle Regular Expression

For MS SQL Server

select distinct city from station where city not like '[aeuio]%';

answered Nov 30 '18 at 4:22

Arun Solomon

112

add a comment |

Mysql we can use this.

select distinct city from station where city regexp '^[^aeiou].*';

To Know More About MySQL Regular Expression

For Oracle we can use

select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');

To Know More About Oracle Regular Expression

For MS SQL Server

select distinct city from station where city not like '[aeuio]%';

answered Nov 30 '18 at 4:22

Arun Solomon

112

Mysql we can use this.

select distinct city from station where city regexp '^[^aeiou].*';

To Know More About MySQL Regular Expression

For Oracle we can use

select distinct(city) from station where regexp_like (city, '^[^AEIOU](*)');

To Know More About Oracle Regular Expression

For MS SQL Server

select distinct city from station where city not like '[aeuio]%';

answered Nov 30 '18 at 4:22

Arun Solomon

112

answered Nov 30 '18 at 4:22

Arun Solomon

112

answered Nov 30 '18 at 4:22

Arun Solomon

112

answered Nov 30 '18 at 4:22

Arun Solomon

112

add a comment |

-1

select distinct city from station where

city not in

(select distinct city from station where

 (city like "A%" or

city like "E%" or

 city like "I%" or

 city like "O%" or

 city like "U%"

));

This will work i guess.

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

answered Oct 18 '16 at 3:04

Saikrishna Syamala

please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27

add a comment |

-1

select distinct city from station where

city not in

(select distinct city from station where

 (city like "A%" or

city like "E%" or

 city like "I%" or

 city like "O%" or

 city like "U%"

));

This will work i guess.

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

answered Oct 18 '16 at 3:04

Saikrishna Syamala

please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27

add a comment |

-1

select distinct city from station where

city not in

(select distinct city from station where

 (city like "A%" or

city like "E%" or

 city like "I%" or

 city like "O%" or

 city like "U%"

));

This will work i guess.

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

answered Oct 18 '16 at 3:04

Saikrishna Syamala

select distinct city from station where

city not in

(select distinct city from station where

 (city like "A%" or

city like "E%" or

 city like "I%" or

 city like "O%" or

 city like "U%"

));

This will work i guess.

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

answered Oct 18 '16 at 3:04

Saikrishna Syamala

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

edited Oct 18 '16 at 4:14

noufalcep

2,25982136

answered Oct 18 '16 at 3:04

Saikrishna Syamala

answered Oct 18 '16 at 3:04

Saikrishna Syamala

answered Oct 18 '16 at 3:04

Saikrishna Syamala

please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27

add a comment |

please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27

please provide justification, why it will work.
– piyushj
Oct 18 '16 at 3:27

add a comment |

-1

SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'

This works for MYSQL

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11

Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19

add a comment |

-1

SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'

This works for MYSQL

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11

Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19

add a comment |

-1

SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'

This works for MYSQL

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

SELECT DISTINCT CITY
FROM STATION
WHERE CITY NOT RLIKE '[aeiouAEIOU]$'

This works for MYSQL

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

answered Dec 9 '17 at 7:33

Vivek Ghanchi

236

You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11

Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19

add a comment |

You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11

Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19

You've anchored the pattern to the end of the string, not the start. This gets you cities that DON'T end in a vowel, the opposite of what was asked. Try
– Paul Campbell
Dec 9 '17 at 11:11

Try SELECT 'Aberdeen' NOT RLIKE '[aeiouAEIOU]$' and see what you get.
– Paul Campbell
Dec 9 '17 at 11:19

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu