How can I eliminate duplicate data in multiple columns query












1














I just asked the question about how I eliminate duplicate data in a column



How can I eliminate duplicate data in column



this code below can delete duplicates in a column



with data as
(
select 'apple, apple, apple, apple' col from dual
)
select listagg(col, ',') within group(order by 1) col
from (
select distinct regexp_substr(col, '[^,]+', 1, level) col
from data
connect by level <= regexp_count(col, ',')
)


next question is
now I do not know how to eliminate data in multiple columns



select 'apple, apple, apple' as col1, 
'prince,prince,princess' as col2,
'dog, cat, cat' as col3
from dual;


I would like to show



COL1     COL2                COL3
----- ---------------- --------
apple prince, princess dog, cat









share|improve this question




















  • 1




    Why do you want to do this exercise and more importantly why in the world do you store all these delimited strings in single columns? It is a bad design practice and should be avoided at all cost. You will get a solution from someone here no doubt, but it is not worth putting such a dreadful structure in production code. The better approach should be to revamp the table structure / schema by following the rules of normalisation.
    – Kaushik Nayak
    Nov 19 '18 at 16:11


















1














I just asked the question about how I eliminate duplicate data in a column



How can I eliminate duplicate data in column



this code below can delete duplicates in a column



with data as
(
select 'apple, apple, apple, apple' col from dual
)
select listagg(col, ',') within group(order by 1) col
from (
select distinct regexp_substr(col, '[^,]+', 1, level) col
from data
connect by level <= regexp_count(col, ',')
)


next question is
now I do not know how to eliminate data in multiple columns



select 'apple, apple, apple' as col1, 
'prince,prince,princess' as col2,
'dog, cat, cat' as col3
from dual;


I would like to show



COL1     COL2                COL3
----- ---------------- --------
apple prince, princess dog, cat









share|improve this question




















  • 1




    Why do you want to do this exercise and more importantly why in the world do you store all these delimited strings in single columns? It is a bad design practice and should be avoided at all cost. You will get a solution from someone here no doubt, but it is not worth putting such a dreadful structure in production code. The better approach should be to revamp the table structure / schema by following the rules of normalisation.
    – Kaushik Nayak
    Nov 19 '18 at 16:11
















1












1








1







I just asked the question about how I eliminate duplicate data in a column



How can I eliminate duplicate data in column



this code below can delete duplicates in a column



with data as
(
select 'apple, apple, apple, apple' col from dual
)
select listagg(col, ',') within group(order by 1) col
from (
select distinct regexp_substr(col, '[^,]+', 1, level) col
from data
connect by level <= regexp_count(col, ',')
)


next question is
now I do not know how to eliminate data in multiple columns



select 'apple, apple, apple' as col1, 
'prince,prince,princess' as col2,
'dog, cat, cat' as col3
from dual;


I would like to show



COL1     COL2                COL3
----- ---------------- --------
apple prince, princess dog, cat









share|improve this question















I just asked the question about how I eliminate duplicate data in a column



How can I eliminate duplicate data in column



this code below can delete duplicates in a column



with data as
(
select 'apple, apple, apple, apple' col from dual
)
select listagg(col, ',') within group(order by 1) col
from (
select distinct regexp_substr(col, '[^,]+', 1, level) col
from data
connect by level <= regexp_count(col, ',')
)


next question is
now I do not know how to eliminate data in multiple columns



select 'apple, apple, apple' as col1, 
'prince,prince,princess' as col2,
'dog, cat, cat' as col3
from dual;


I would like to show



COL1     COL2                COL3
----- ---------------- --------
apple prince, princess dog, cat






sql oracle






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 19 '18 at 16:54









Barbaros Özhan

12.3k71530




12.3k71530










asked Nov 19 '18 at 15:54









supercool djkazu

5261410




5261410








  • 1




    Why do you want to do this exercise and more importantly why in the world do you store all these delimited strings in single columns? It is a bad design practice and should be avoided at all cost. You will get a solution from someone here no doubt, but it is not worth putting such a dreadful structure in production code. The better approach should be to revamp the table structure / schema by following the rules of normalisation.
    – Kaushik Nayak
    Nov 19 '18 at 16:11
















  • 1




    Why do you want to do this exercise and more importantly why in the world do you store all these delimited strings in single columns? It is a bad design practice and should be avoided at all cost. You will get a solution from someone here no doubt, but it is not worth putting such a dreadful structure in production code. The better approach should be to revamp the table structure / schema by following the rules of normalisation.
    – Kaushik Nayak
    Nov 19 '18 at 16:11










1




1




Why do you want to do this exercise and more importantly why in the world do you store all these delimited strings in single columns? It is a bad design practice and should be avoided at all cost. You will get a solution from someone here no doubt, but it is not worth putting such a dreadful structure in production code. The better approach should be to revamp the table structure / schema by following the rules of normalisation.
– Kaushik Nayak
Nov 19 '18 at 16:11






Why do you want to do this exercise and more importantly why in the world do you store all these delimited strings in single columns? It is a bad design practice and should be avoided at all cost. You will get a solution from someone here no doubt, but it is not worth putting such a dreadful structure in production code. The better approach should be to revamp the table structure / schema by following the rules of normalisation.
– Kaushik Nayak
Nov 19 '18 at 16:11














1 Answer
1






active

oldest

votes


















1














You may use such a combination :



select  
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('apple, apple, apple','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('apple, apple, apple',',') + 1
)
) as str1,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('prince,prince,princess','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('prince,prince,princess',',') + 1
)
) as str2,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('dog, cat, cat','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('dog, cat, cat',',') + 1
)
) as str3
from dual;

STR1 STR2 STR3
------ --------------- --------
apple prince,princess cat,dog


Rextester Demo






share|improve this answer





















  • Thanks you so much!!!
    – supercool djkazu
    Nov 19 '18 at 16:38










  • @supercooldjkazu you're very welcome friend.
    – Barbaros Özhan
    Nov 19 '18 at 16:51











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You may use such a combination :



select  
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('apple, apple, apple','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('apple, apple, apple',',') + 1
)
) as str1,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('prince,prince,princess','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('prince,prince,princess',',') + 1
)
) as str2,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('dog, cat, cat','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('dog, cat, cat',',') + 1
)
) as str3
from dual;

STR1 STR2 STR3
------ --------------- --------
apple prince,princess cat,dog


Rextester Demo






share|improve this answer





















  • Thanks you so much!!!
    – supercool djkazu
    Nov 19 '18 at 16:38










  • @supercooldjkazu you're very welcome friend.
    – Barbaros Özhan
    Nov 19 '18 at 16:51
















1














You may use such a combination :



select  
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('apple, apple, apple','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('apple, apple, apple',',') + 1
)
) as str1,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('prince,prince,princess','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('prince,prince,princess',',') + 1
)
) as str2,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('dog, cat, cat','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('dog, cat, cat',',') + 1
)
) as str3
from dual;

STR1 STR2 STR3
------ --------------- --------
apple prince,princess cat,dog


Rextester Demo






share|improve this answer





















  • Thanks you so much!!!
    – supercool djkazu
    Nov 19 '18 at 16:38










  • @supercooldjkazu you're very welcome friend.
    – Barbaros Özhan
    Nov 19 '18 at 16:51














1












1








1






You may use such a combination :



select  
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('apple, apple, apple','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('apple, apple, apple',',') + 1
)
) as str1,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('prince,prince,princess','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('prince,prince,princess',',') + 1
)
) as str2,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('dog, cat, cat','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('dog, cat, cat',',') + 1
)
) as str3
from dual;

STR1 STR2 STR3
------ --------------- --------
apple prince,princess cat,dog


Rextester Demo






share|improve this answer












You may use such a combination :



select  
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('apple, apple, apple','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('apple, apple, apple',',') + 1
)
) as str1,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('prince,prince,princess','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('prince,prince,princess',',') + 1
)
) as str2,
(
select listagg(str,',') within group (order by 0)
from
(
select distinct trim(regexp_substr('dog, cat, cat','[^,]+', 1, level)) as str
from dual
connect by level <= regexp_count('dog, cat, cat',',') + 1
)
) as str3
from dual;

STR1 STR2 STR3
------ --------------- --------
apple prince,princess cat,dog


Rextester Demo







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 19 '18 at 16:25









Barbaros Özhan

12.3k71530




12.3k71530












  • Thanks you so much!!!
    – supercool djkazu
    Nov 19 '18 at 16:38










  • @supercooldjkazu you're very welcome friend.
    – Barbaros Özhan
    Nov 19 '18 at 16:51


















  • Thanks you so much!!!
    – supercool djkazu
    Nov 19 '18 at 16:38










  • @supercooldjkazu you're very welcome friend.
    – Barbaros Özhan
    Nov 19 '18 at 16:51
















Thanks you so much!!!
– supercool djkazu
Nov 19 '18 at 16:38




Thanks you so much!!!
– supercool djkazu
Nov 19 '18 at 16:38












@supercooldjkazu you're very welcome friend.
– Barbaros Özhan
Nov 19 '18 at 16:51




@supercooldjkazu you're very welcome friend.
– Barbaros Özhan
Nov 19 '18 at 16:51


















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