How can I output the number on the nth position in a string?

-2

Using bash, lets say i have the following string

string="Same bought 5 bananas, 12 apples, 2 peaches and 16 oranges"

How can I trim everything except the nth number. In this case I want to output 12 which is the second number in the string.

How can I do that with bash, grep or sed?

edited Jan 16 at 12:12

asked Dec 12 '18 at 11:28

Bret Joseph

8410

If X is unknown, how do you establish that it's X?

– tripleee
Dec 12 '18 at 11:32

@tripleee im representing the unknown with X.. X is actually an interger

– Bret Joseph
Dec 12 '18 at 11:34

What I am trying to say is that your question is unclear. What would an acceptable answer look like if we don't know what X is, or how we can find out?

– tripleee
Dec 12 '18 at 11:39

Obviously if you know that X is 1, the answer is trivial; sed 's/.*1.*/1/ but then what do you need the variable for? The result after the substitution is clearly identical to the input string, so there is no point in performing any substitution. Or are you trying to find out if the string contains 1? case $string in *1*) echo true;; esac

– tripleee
Dec 12 '18 at 11:40

the problem is I don't want to use the character 1 since X it not always 1

– Bret Joseph
Dec 12 '18 at 11:44

|
show 3 more comments

-2

Using bash, lets say i have the following string

string="Same bought 5 bananas, 12 apples, 2 peaches and 16 oranges"

How can I trim everything except the nth number. In this case I want to output 12 which is the second number in the string.

How can I do that with bash, grep or sed?

edited Jan 16 at 12:12

asked Dec 12 '18 at 11:28

Bret Joseph

8410

If X is unknown, how do you establish that it's X?

– tripleee
Dec 12 '18 at 11:32

@tripleee im representing the unknown with X.. X is actually an interger

– Bret Joseph
Dec 12 '18 at 11:34

What I am trying to say is that your question is unclear. What would an acceptable answer look like if we don't know what X is, or how we can find out?

– tripleee
Dec 12 '18 at 11:39

Obviously if you know that X is 1, the answer is trivial; sed 's/.*1.*/1/ but then what do you need the variable for? The result after the substitution is clearly identical to the input string, so there is no point in performing any substitution. Or are you trying to find out if the string contains 1? case $string in *1*) echo true;; esac

– tripleee
Dec 12 '18 at 11:40

the problem is I don't want to use the character 1 since X it not always 1

– Bret Joseph
Dec 12 '18 at 11:44

|
show 3 more comments

-2

Using bash, lets say i have the following string

string="Same bought 5 bananas, 12 apples, 2 peaches and 16 oranges"

How can I trim everything except the nth number. In this case I want to output 12 which is the second number in the string.

How can I do that with bash, grep or sed?

edited Jan 16 at 12:12

asked Dec 12 '18 at 11:28

Bret Joseph

8410

Using bash, lets say i have the following string

string="Same bought 5 bananas, 12 apples, 2 peaches and 16 oranges"

How can I trim everything except the nth number. In this case I want to output 12 which is the second number in the string.

How can I do that with bash, grep or sed?

bash sed grep text-manipulation

edited Jan 16 at 12:12

asked Dec 12 '18 at 11:28

Bret Joseph

8410

edited Jan 16 at 12:12

asked Dec 12 '18 at 11:28

Bret Joseph

8410

edited Jan 16 at 12:12

asked Dec 12 '18 at 11:28

Bret Joseph

8410

asked Dec 12 '18 at 11:28

Bret Joseph

8410

asked Dec 12 '18 at 11:28

Bret Joseph

8410

If X is unknown, how do you establish that it's X?

– tripleee
Dec 12 '18 at 11:32

@tripleee im representing the unknown with X.. X is actually an interger

– Bret Joseph
Dec 12 '18 at 11:34

What I am trying to say is that your question is unclear. What would an acceptable answer look like if we don't know what X is, or how we can find out?

– tripleee
Dec 12 '18 at 11:39

Obviously if you know that X is 1, the answer is trivial; sed 's/.*1.*/1/ but then what do you need the variable for? The result after the substitution is clearly identical to the input string, so there is no point in performing any substitution. Or are you trying to find out if the string contains 1? case $string in *1*) echo true;; esac

– tripleee
Dec 12 '18 at 11:40

the problem is I don't want to use the character 1 since X it not always 1

– Bret Joseph
Dec 12 '18 at 11:44

|
show 3 more comments

If X is unknown, how do you establish that it's X?

– tripleee
Dec 12 '18 at 11:32

@tripleee im representing the unknown with X.. X is actually an interger

– Bret Joseph
Dec 12 '18 at 11:34

What I am trying to say is that your question is unclear. What would an acceptable answer look like if we don't know what X is, or how we can find out?

– tripleee
Dec 12 '18 at 11:39

Obviously if you know that X is 1, the answer is trivial; sed 's/.*1.*/1/ but then what do you need the variable for? The result after the substitution is clearly identical to the input string, so there is no point in performing any substitution. Or are you trying to find out if the string contains 1? case $string in *1*) echo true;; esac

– tripleee
Dec 12 '18 at 11:40

the problem is I don't want to use the character 1 since X it not always 1

– Bret Joseph
Dec 12 '18 at 11:44

If X is unknown, how do you establish that it's X?

– tripleee
Dec 12 '18 at 11:32

@tripleee im representing the unknown with X.. X is actually an interger

– Bret Joseph
Dec 12 '18 at 11:34

What I am trying to say is that your question is unclear. What would an acceptable answer look like if we don't know what X is, or how we can find out?

– tripleee
Dec 12 '18 at 11:39

Obviously if you know that X is 1, the answer is trivial; sed 's/.*1.*/1/ but then what do you need the variable for? The result after the substitution is clearly identical to the input string, so there is no point in performing any substitution. Or are you trying to find out if the string contains 1? case $string in *1*) echo true;; esac

– tripleee
Dec 12 '18 at 11:40

the problem is I don't want to use the character 1 since X it not always 1

– Bret Joseph
Dec 12 '18 at 11:44

|
show 3 more comments

2 Answers
2

active

oldest

votes

This might work for you (GNU sed):

sed 's/^([^0-9]*([0-9]*)){2}.*/2/;/^$/d' file

This replaces the current line by the second occurrence of a group of numbers. The line is deleted unless a number is output.

answered Dec 12 '18 at 14:33

potong

36k43062

thanks for the effort, what do you think of my own solution as well

– Bret Joseph
Jan 1 at 5:16

add a comment |

Solution Using sed

According to Paul Hodges and Triplee
grep -Eo '[0-9]+' <<<"$string"| sed 'nq;d'

Where n is the position of the number

sed 'NUMq;d'
NUM is the lines to print.
2q says quit on the second line.
d will delete every other line except the last

edited Jan 2 at 17:14

answered Dec 12 '18 at 12:25

Bret Joseph

8410

3

'2q' says quit on the second line, but won't stop the line from being printed. 'd' says delete every line you come to, which does prevent those lines from being printed. It checks them in order though, so '2q;d' will quit (and flush-print) on line 2 before it hits the delete, but not on line one, so the d prevents printing of line 1 but not line 2.

– Paul Hodges
Dec 12 '18 at 15:07

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

This might work for you (GNU sed):

sed 's/^([^0-9]*([0-9]*)){2}.*/2/;/^$/d' file

This replaces the current line by the second occurrence of a group of numbers. The line is deleted unless a number is output.

answered Dec 12 '18 at 14:33

potong

36k43062

thanks for the effort, what do you think of my own solution as well

– Bret Joseph
Jan 1 at 5:16

add a comment |

This might work for you (GNU sed):

sed 's/^([^0-9]*([0-9]*)){2}.*/2/;/^$/d' file

This replaces the current line by the second occurrence of a group of numbers. The line is deleted unless a number is output.

answered Dec 12 '18 at 14:33

potong

36k43062

thanks for the effort, what do you think of my own solution as well

– Bret Joseph
Jan 1 at 5:16

add a comment |

This might work for you (GNU sed):

sed 's/^([^0-9]*([0-9]*)){2}.*/2/;/^$/d' file

This replaces the current line by the second occurrence of a group of numbers. The line is deleted unless a number is output.

answered Dec 12 '18 at 14:33

potong

36k43062

This might work for you (GNU sed):

sed 's/^([^0-9]*([0-9]*)){2}.*/2/;/^$/d' file

This replaces the current line by the second occurrence of a group of numbers. The line is deleted unless a number is output.

answered Dec 12 '18 at 14:33

potong

36k43062

answered Dec 12 '18 at 14:33

potong

36k43062

answered Dec 12 '18 at 14:33

potong

36k43062

answered Dec 12 '18 at 14:33

potong

36k43062

thanks for the effort, what do you think of my own solution as well

– Bret Joseph
Jan 1 at 5:16

add a comment |

thanks for the effort, what do you think of my own solution as well

– Bret Joseph
Jan 1 at 5:16

thanks for the effort, what do you think of my own solution as well

– Bret Joseph
Jan 1 at 5:16

add a comment |

Solution Using sed

According to Paul Hodges and Triplee
grep -Eo '[0-9]+' <<<"$string"| sed 'nq;d'

Where n is the position of the number

sed 'NUMq;d'
NUM is the lines to print.
2q says quit on the second line.
d will delete every other line except the last

edited Jan 2 at 17:14

answered Dec 12 '18 at 12:25

Bret Joseph

8410

3

'2q' says quit on the second line, but won't stop the line from being printed. 'd' says delete every line you come to, which does prevent those lines from being printed. It checks them in order though, so '2q;d' will quit (and flush-print) on line 2 before it hits the delete, but not on line one, so the d prevents printing of line 1 but not line 2.

– Paul Hodges
Dec 12 '18 at 15:07

add a comment |

Solution Using sed

According to Paul Hodges and Triplee
grep -Eo '[0-9]+' <<<"$string"| sed 'nq;d'

Where n is the position of the number

sed 'NUMq;d'
NUM is the lines to print.
2q says quit on the second line.
d will delete every other line except the last

edited Jan 2 at 17:14

answered Dec 12 '18 at 12:25

Bret Joseph

8410

3

'2q' says quit on the second line, but won't stop the line from being printed. 'd' says delete every line you come to, which does prevent those lines from being printed. It checks them in order though, so '2q;d' will quit (and flush-print) on line 2 before it hits the delete, but not on line one, so the d prevents printing of line 1 but not line 2.

– Paul Hodges
Dec 12 '18 at 15:07

add a comment |

Solution Using sed

According to Paul Hodges and Triplee
grep -Eo '[0-9]+' <<<"$string"| sed 'nq;d'

Where n is the position of the number

sed 'NUMq;d'
NUM is the lines to print.
2q says quit on the second line.
d will delete every other line except the last

edited Jan 2 at 17:14

answered Dec 12 '18 at 12:25

Bret Joseph

8410

Solution Using sed

According to Paul Hodges and Triplee
grep -Eo '[0-9]+' <<<"$string"| sed 'nq;d'

Where n is the position of the number

sed 'NUMq;d'
NUM is the lines to print.
2q says quit on the second line.
d will delete every other line except the last

edited Jan 2 at 17:14

answered Dec 12 '18 at 12:25

Bret Joseph

8410

edited Jan 2 at 17:14

answered Dec 12 '18 at 12:25

Bret Joseph

8410

answered Dec 12 '18 at 12:25

Bret Joseph

8410

answered Dec 12 '18 at 12:25

Bret Joseph

8410

3

'2q' says quit on the second line, but won't stop the line from being printed. 'd' says delete every line you come to, which does prevent those lines from being printed. It checks them in order though, so '2q;d' will quit (and flush-print) on line 2 before it hits the delete, but not on line one, so the d prevents printing of line 1 but not line 2.

– Paul Hodges
Dec 12 '18 at 15:07

add a comment |

3

'2q' says quit on the second line, but won't stop the line from being printed. 'd' says delete every line you come to, which does prevent those lines from being printed. It checks them in order though, so '2q;d' will quit (and flush-print) on line 2 before it hits the delete, but not on line one, so the d prevents printing of line 1 but not line 2.

– Paul Hodges
Dec 12 '18 at 15:07

'2q' says quit on the second line, but won't stop the line from being printed. 'd' says delete every line you come to, which does prevent those lines from being printed. It checks them in order though, so '2q;d' will quit (and flush-print) on line 2 before it hits the delete, but not on line one, so the d prevents printing of line 1 but not line 2.

– Paul Hodges
Dec 12 '18 at 15:07

add a comment |

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