Can not extract the capture group with neither sed nor grep

I want to extract the value pair from a key-value pair syntax but I can not.

Example I tried:

echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'

But this gives employee_id=1234 and not 1234 which is actually the capture group.

What am I doing wrong here? I also tried:

echo employee_id=1234| egrep -o employee_id=([0-9]+)

but no sucess.

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

asked Sep 19 '13 at 10:52

Jim

7,4862395179

1

AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.

– Panos Rontogiannis
Sep 19 '13 at 11:50

1

possible duplicate of How to escape plus sign on mac os x (BSD) sed?

– Michael Foukarakis
Sep 19 '13 at 14:12

echo 'employee_id=1234' | cut -d '=' -f 2

– ALex_hha
Mar 11 '16 at 15:53

add a comment |

I want to extract the value pair from a key-value pair syntax but I can not.

Example I tried:

echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'

But this gives employee_id=1234 and not 1234 which is actually the capture group.

What am I doing wrong here? I also tried:

echo employee_id=1234| egrep -o employee_id=([0-9]+)

but no sucess.

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

asked Sep 19 '13 at 10:52

Jim

7,4862395179

1

AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.

– Panos Rontogiannis
Sep 19 '13 at 11:50

1

possible duplicate of How to escape plus sign on mac os x (BSD) sed?

– Michael Foukarakis
Sep 19 '13 at 14:12

echo 'employee_id=1234' | cut -d '=' -f 2

– ALex_hha
Mar 11 '16 at 15:53

add a comment |

I want to extract the value pair from a key-value pair syntax but I can not.

Example I tried:

echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'

But this gives employee_id=1234 and not 1234 which is actually the capture group.

What am I doing wrong here? I also tried:

echo employee_id=1234| egrep -o employee_id=([0-9]+)

but no sucess.

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

asked Sep 19 '13 at 10:52

Jim

7,4862395179

I want to extract the value pair from a key-value pair syntax but I can not.

Example I tried:

echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'

But this gives employee_id=1234 and not 1234 which is actually the capture group.

What am I doing wrong here? I also tried:

echo employee_id=1234| egrep -o employee_id=([0-9]+)

but no sucess.

regex linux sed grep

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

asked Sep 19 '13 at 10:52

Jim

7,4862395179

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

asked Sep 19 '13 at 10:52

Jim

7,4862395179

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

edited Sep 19 '13 at 18:41

Adrian Frühwirth

26k74963

asked Sep 19 '13 at 10:52

Jim

7,4862395179

asked Sep 19 '13 at 10:52

Jim

7,4862395179

asked Sep 19 '13 at 10:52

Jim

7,4862395179

1

AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.

– Panos Rontogiannis
Sep 19 '13 at 11:50

1

possible duplicate of How to escape plus sign on mac os x (BSD) sed?

– Michael Foukarakis
Sep 19 '13 at 14:12

echo 'employee_id=1234' | cut -d '=' -f 2

– ALex_hha
Mar 11 '16 at 15:53

add a comment |

1

AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.

– Panos Rontogiannis
Sep 19 '13 at 11:50

1

possible duplicate of How to escape plus sign on mac os x (BSD) sed?

– Michael Foukarakis
Sep 19 '13 at 14:12

echo 'employee_id=1234' | cut -d '=' -f 2

– ALex_hha
Mar 11 '16 at 15:53

AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.

– Panos Rontogiannis
Sep 19 '13 at 11:50

possible duplicate of How to escape plus sign on mac os x (BSD) sed?

– Michael Foukarakis
Sep 19 '13 at 14:12

echo 'employee_id=1234' | cut -d '=' -f 2

– ALex_hha
Mar 11 '16 at 15:53

add a comment |

5 Answers
5

active

oldest

votes

1. Use egrep -o:

echo 'employee_id=1234' | egrep -o '[0-9]+'

1234

2. using grep -oP (PCRE):

echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'

1234

3. Using sed:

echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'

1234

edited Sep 19 '13 at 11:20

answered Sep 19 '13 at 10:56

anubhava

526k46324398

+1. But why does mine doesn't work. I used egrep -o

– Jim
Sep 19 '13 at 10:58

Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after employee_id=

– Jim
Sep 19 '13 at 11:00

The last command in your comment does not work at all

– Jim
Sep 19 '13 at 11:01

1

None of your answers are relevant. The first I can not use because I need only the number after employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?

– Jim
Sep 19 '13 at 11:06

1

2 worked for me as a way to get PHP version for use in an Ansible playbook php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"

– turrican_34
Dec 18 '18 at 14:04

|
show 5 more comments

To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:

$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"

$ echo $some_string | grep -oP "$regex"

# matches and returns b

$ echo "abc" | grep -oP "aK(b)(?=c)" 

b 

# no match

$ echo "abc" | grep -oP "zK(b)(?=c)"

# no match

$ echo "abc" | grep -oP "aK(b)(?=d)"

edited May 12 '16 at 22:25

answered Apr 6 '16 at 15:34

jayflo

5401619

add a comment |

Using awk

echo 'employee_id=1234' | awk -F= '{print $2}'

1234

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

add a comment |

You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:

foo='employee_id=1234'

var=${foo%%=*}

value=${foo#*=}

$ echo "var=${var} value=${value}"

var=employee_id value=1234

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

add a comment |

use sed -E for extended regex

    echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'

answered Nov 20 '18 at 5:34

commander Ghost

111

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

1. Use egrep -o:

echo 'employee_id=1234' | egrep -o '[0-9]+'

1234

2. using grep -oP (PCRE):

echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'

1234

3. Using sed:

echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'

1234

edited Sep 19 '13 at 11:20

answered Sep 19 '13 at 10:56

anubhava

526k46324398

+1. But why does mine doesn't work. I used egrep -o

– Jim
Sep 19 '13 at 10:58

Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after employee_id=

– Jim
Sep 19 '13 at 11:00

The last command in your comment does not work at all

– Jim
Sep 19 '13 at 11:01

1

None of your answers are relevant. The first I can not use because I need only the number after employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?

– Jim
Sep 19 '13 at 11:06

1

2 worked for me as a way to get PHP version for use in an Ansible playbook php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"

– turrican_34
Dec 18 '18 at 14:04

|
show 5 more comments

1. Use egrep -o:

echo 'employee_id=1234' | egrep -o '[0-9]+'

1234

2. using grep -oP (PCRE):

echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'

1234

3. Using sed:

echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'

1234

edited Sep 19 '13 at 11:20

answered Sep 19 '13 at 10:56

anubhava

526k46324398

+1. But why does mine doesn't work. I used egrep -o

– Jim
Sep 19 '13 at 10:58

Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after employee_id=

– Jim
Sep 19 '13 at 11:00

The last command in your comment does not work at all

– Jim
Sep 19 '13 at 11:01

1

None of your answers are relevant. The first I can not use because I need only the number after employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?

– Jim
Sep 19 '13 at 11:06

1

2 worked for me as a way to get PHP version for use in an Ansible playbook php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"

– turrican_34
Dec 18 '18 at 14:04

|
show 5 more comments

1. Use egrep -o:

echo 'employee_id=1234' | egrep -o '[0-9]+'

1234

2. using grep -oP (PCRE):

echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'

1234

3. Using sed:

echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'

1234

edited Sep 19 '13 at 11:20

answered Sep 19 '13 at 10:56

anubhava

526k46324398

1. Use egrep -o:

echo 'employee_id=1234' | egrep -o '[0-9]+'

1234

2. using grep -oP (PCRE):

echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'

1234

3. Using sed:

echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'

1234

edited Sep 19 '13 at 11:20

answered Sep 19 '13 at 10:56

anubhava

526k46324398

edited Sep 19 '13 at 11:20

answered Sep 19 '13 at 10:56

anubhava

526k46324398

answered Sep 19 '13 at 10:56

anubhava

526k46324398

answered Sep 19 '13 at 10:56

anubhava

526k46324398

+1. But why does mine doesn't work. I used egrep -o

– Jim
Sep 19 '13 at 10:58

Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after employee_id=

– Jim
Sep 19 '13 at 11:00

The last command in your comment does not work at all

– Jim
Sep 19 '13 at 11:01

1

None of your answers are relevant. The first I can not use because I need only the number after employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?

– Jim
Sep 19 '13 at 11:06

1

2 worked for me as a way to get PHP version for use in an Ansible playbook php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"

– turrican_34
Dec 18 '18 at 14:04

|
show 5 more comments

+1. But why does mine doesn't work. I used egrep -o

– Jim
Sep 19 '13 at 10:58

Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after employee_id=

– Jim
Sep 19 '13 at 11:00

The last command in your comment does not work at all

– Jim
Sep 19 '13 at 11:01

1

None of your answers are relevant. The first I can not use because I need only the number after employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?

– Jim
Sep 19 '13 at 11:06

1

2 worked for me as a way to get PHP version for use in an Ansible playbook php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"

– turrican_34
Dec 18 '18 at 14:04

+1. But why does mine doesn't work. I used egrep -o

– Jim
Sep 19 '13 at 10:58

Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after employee_id=

– Jim
Sep 19 '13 at 11:00

The last command in your comment does not work at all

– Jim
Sep 19 '13 at 11:01

None of your answers are relevant. The first I can not use because I need only the number after employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?

– Jim
Sep 19 '13 at 11:06

2 worked for me as a way to get PHP version for use in an Ansible playbook php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"

– turrican_34
Dec 18 '18 at 14:04

|
show 5 more comments

To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:

$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"

$ echo $some_string | grep -oP "$regex"

# matches and returns b

$ echo "abc" | grep -oP "aK(b)(?=c)" 

b 

# no match

$ echo "abc" | grep -oP "zK(b)(?=c)"

# no match

$ echo "abc" | grep -oP "aK(b)(?=d)"

edited May 12 '16 at 22:25

answered Apr 6 '16 at 15:34

jayflo

5401619

add a comment |

To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:

$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"

$ echo $some_string | grep -oP "$regex"

# matches and returns b

$ echo "abc" | grep -oP "aK(b)(?=c)" 

b 

# no match

$ echo "abc" | grep -oP "zK(b)(?=c)"

# no match

$ echo "abc" | grep -oP "aK(b)(?=d)"

edited May 12 '16 at 22:25

answered Apr 6 '16 at 15:34

jayflo

5401619

add a comment |

To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:

$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"

$ echo $some_string | grep -oP "$regex"

# matches and returns b

$ echo "abc" | grep -oP "aK(b)(?=c)" 

b 

# no match

$ echo "abc" | grep -oP "zK(b)(?=c)"

# no match

$ echo "abc" | grep -oP "aK(b)(?=d)"

edited May 12 '16 at 22:25

answered Apr 6 '16 at 15:34

jayflo

5401619

To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:

$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"

$ echo $some_string | grep -oP "$regex"

# matches and returns b

$ echo "abc" | grep -oP "aK(b)(?=c)" 

b 

# no match

$ echo "abc" | grep -oP "zK(b)(?=c)"

# no match

$ echo "abc" | grep -oP "aK(b)(?=d)"

edited May 12 '16 at 22:25

answered Apr 6 '16 at 15:34

jayflo

5401619

edited May 12 '16 at 22:25

answered Apr 6 '16 at 15:34

jayflo

5401619

answered Apr 6 '16 at 15:34

jayflo

5401619

answered Apr 6 '16 at 15:34

jayflo

5401619

add a comment |

Using awk

echo 'employee_id=1234' | awk -F= '{print $2}'

1234

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

add a comment |

Using awk

echo 'employee_id=1234' | awk -F= '{print $2}'

1234

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

add a comment |

Using awk

echo 'employee_id=1234' | awk -F= '{print $2}'

1234

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

Using awk

echo 'employee_id=1234' | awk -F= '{print $2}'

1234

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

answered Sep 19 '13 at 11:29

Jotne

29.5k73245

add a comment |

You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:

foo='employee_id=1234'

var=${foo%%=*}

value=${foo#*=}

$ echo "var=${var} value=${value}"

var=employee_id value=1234

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

add a comment |

You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:

foo='employee_id=1234'

var=${foo%%=*}

value=${foo#*=}

$ echo "var=${var} value=${value}"

var=employee_id value=1234

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

add a comment |

You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:

foo='employee_id=1234'

var=${foo%%=*}

value=${foo#*=}

$ echo "var=${var} value=${value}"

var=employee_id value=1234

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:

foo='employee_id=1234'

var=${foo%%=*}

value=${foo#*=}

$ echo "var=${var} value=${value}"

var=employee_id value=1234

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

answered Sep 19 '13 at 14:00

Adrian Frühwirth

26k74963

add a comment |

use sed -E for extended regex

    echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'

answered Nov 20 '18 at 5:34

commander Ghost

111

add a comment |

use sed -E for extended regex

    echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'

answered Nov 20 '18 at 5:34

commander Ghost

111

add a comment |

use sed -E for extended regex

    echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'

answered Nov 20 '18 at 5:34

commander Ghost

111

use sed -E for extended regex

    echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'

answered Nov 20 '18 at 5:34

commander Ghost

111

answered Nov 20 '18 at 5:34

commander Ghost

111

answered Nov 20 '18 at 5:34

commander Ghost

111

answered Nov 20 '18 at 5:34

commander Ghost

111

add a comment |

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