How to output only captured groups with sed

Question

Is there any way to tell sed to output only captured groups  For example given the input   This is a sample 123 text and some 987 numbers   and pattern       d       Could I get only 123 and 987 output in the way formatted by back references

User · Answer

run s  of digits This answer works with any count of digit groups  Example    echo  Num123that456are7899900contained0018166intext         sed -En  s   0-9    0-9  1     0-9    1  gp   123 456 7899900 0018166  Expanded answer   Is there any way to tell sed to output only captured groups   Yes  replace all text by the capture group    echo  Number 123 inside text         sed  s   0-9     0-9   1       0-9    1    123  s   0-9                               several non-digits             0-9   1                   followed by one or more digits                           0-9         and followed by more non-digits                                   1    gets replaced only by the digits   Or with extended syntax  less backquotes and allow the use of       echo  Number 123 in text         sed -E  s   0-9    0-9     0-9    1    123  To avoid printing the original text when there is no number  use    echo  Number xxx in text         sed -En  s   0-9    0-9     0-9    1 p     -n  Do not print the input by default    p  print only if a replacement was done   And to match several numbers  and also print them     echo  N 123 in 456 text        sed -En  s   0-9    0-9     0-9    1  gp   123 456  That works for any count of digit runs    str  Test Num s  123 456 7899900 contained as0018166df in text    echo  quot  str quot         sed -En  s   0-9    0-9  1     0-9    1  gp   123 456 7899900 0018166  Which is very similar to the grep command    str  Test Num s  123 456 7899900 contained as0018166df in text    echo  quot  str quot    grep -Po   d   123 456 7899900 0018166  About  d  and pattern      d      Sed does not recognize the   d   shortcut  syntax  The ascii equivalent used above  0-9  is not exactly equivalent  The only alternative solution is to use a character class      digit      The selected answer use such  quot character classes quot  to build a solution    str  This is a sample 123 text and some 987 numbers    echo  quot  str quot    sed -rn  s     digit        digit         digit        digit         digit      1  2 p   That solution only works for  exactly  two runs of digits  Of course  as the answer is being executed inside the shell  we can define a couple of variables to make such answer shorter    str  This is a sample 123 text and some 987 numbers    d    digit        D     digit      echo  quot  str quot    sed -rn  quot s  D   d   D   d   D   1  2 p quot   But  as has been already explained  using a s         gp command is better    str  This is 75577 a sam33ple 123 text and some 987 numbers    d    digit        D     digit      echo  quot  str quot    sed -rn  quot s  D   d   D   1  gp quot  75577 33 123 987  That will cover both repeated runs of digits and writing a short er  command

User · Answer

Try  sed -n -e    0-9  s    0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9         1  2  3  4  5  6  7  8  9 p    I got this under cygwin      echo  asdf        echo  1234        echo  asdf1234adsf1234asdf        echo  1m2m3m4m5m6m7m8m9m0m1m2m3m4m5m6m7m8m9         sed -n -e    0-9  s    0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9      0-9     0-9         1  2  3  4  5  6  7  8  9 p   1234 1234 1234 1 2 3 4 5 6 7 8 9

User · Answer

It s not what the OP asked for  capturing groups  but you can extract the numbers using   S  This is a sample 123 text and some 987 numbers  echo   S    sed  s    n g    sed -r     0-9      d    Gives the following   123 987

User · Answer

Give up and use Perl  Since sed does not cut it  let s just throw the towel and use Perl  at least it is LSB while grep GNU extensions are not  -    Print the entire matching part  no matching groups or lookbehind needed   cat  lt  lt EOS   perl -lane  print m  d  g  a1 b2 a34 b56 EOS   Output   12 3456  Single match per line  often structured data fields   cat  lt  lt EOS   perl -lape  s    a  d      1 g  a1 b2 a34 b56 EOS   Output   1 34   With lookbehind   cat  lt  lt EOS   perl -lane  print m    lt  a   d     a1 b2 a34 b56 EOS  Multiple fields   cat  lt  lt EOS   perl -lape  s    a  d     b  d      1  2 g  a1 c0 b2 c0 a34 c0 b56 c0 EOS   Output   1 2 34 56  Multiple matches per line  often unstructured data   cat  lt  lt EOS   perl -lape  s    a  d       1  g  a1 b2 a34 b56 a78 b90 EOS   Output   1  34 78   With lookbehind   cat EOS lt  lt    perl -lane  print m    lt  a   d   g  a1 b2 a34 b56 a78 b90 EOS   Output   1 3478

User · Answer

Sed has up to nine remembered patterns but you need to use escaped parentheses to remember portions of the regular expression   See here for examples and more detail

User · Answer

you can use grep  grep -Eow   0-9    file

User · Answer

I believe the pattern given in the question was by way of example only  and the goal was to match any pattern    If you have a sed with the GNU extension allowing insertion of a newline in the pattern space  one suggestion is    gt  set string    This is a sample 123 text and some 987 numbers   gt   gt  set pattern     0-9  0-9     gt  echo  string   sed  s  pattern  n amp  n g    sed -n    pattern p  123 987  gt  set pattern     a-z  a-z     gt  echo  string   sed  s  pattern  n amp  n g    sed -n    pattern p  his is a sample text and some numbers   These examples are with tcsh  yes  I know its the wrong shell  with CYGWIN   Edit  For bash  remove set  and the spaces around

User · Answer

The key to getting this to work is to tell sed to exclude what you don t want to be output as well as specifying what you do want   string  This is a sample 123 text and some 987 numbers  echo   string    sed -rn  s     digit        digit         digit        digit         digit      1  2 p    This says    don t default to printing each line  -n  exclude zero or more non-digits include one or more digits exclude one or more non-digits include one or more digits exclude zero or more non-digits print the substitution  p    In general  in sed you capture groups using parentheses and output what you capture using a back reference   echo  foobarbaz    sed  s  foo      baz   1     will output  bar   If you use -r  -E for OS X  for extended regex  you don t need to escape the parentheses   echo  foobarbaz    sed -r  s  foo    baz   1     There can be up to 9 capture groups and their back references  The back references are numbered in the order the groups appear  but they can be used in any order and can be repeated   echo  foobarbaz    sed -r  s  foo    b   z   2  1  2     outputs  a bar a    If you have GNU grep  it may also work in BSD  including OS X    echo   string    grep -Po   d     or variations such as   echo   string    grep -Po     lt   D    d      The -P option enables Perl Compatible Regular Expressions  See man 3 pcrepattern or man  3 pcresyntax

[regex] How to output only captured groups with sed?

run(s) of digits

Expanded answer.

About \d

Examples related to regex

Examples related to sed