How to print matched regex pattern using awk

Question

Using awk  I need to find a word in a file that matches a regex pattern    I only want to print the word matched with the pattern   So if in the line  I have   xxx yyy zzz   And pattern    yyy    I want to only get   yyy     EDIT  thanks to kurumi i managed to write something like this   awk            for i 1  i lt  NF  i                      tmp match  i    0-9        A-Za-z0-9                    if tmp                            print  i                                 1   and this is what i needed    thanks a lot

User · Answer

If Perl is an option  you can try this   perl -lne  print  1 if   regex    file   To implement case-insensitive matching  add the i modifier  perl -lne  print  1 if   regex  i  file   To print everything AFTER the match   perl -lne  if   found  print  else if   regex       print  1   found      textfile   To print the match and everything after the match   perl -lne  if   found  print  else if    regex      print  1   found      textfile

User · Answer

If you know what column the text pattern you re looking for  e g   yyy   is in  you can just check that specific column to see if it matches  and print it   For example  given a file with the following contents   called asdf txt   xxx yyy zzz   to only print the second column if it matches the pattern  yyy   you could do something like this   awk   2    yyy   print  2   asdf txt   Note that this will also match basically any line where the second column has a  yyy  in it  like these   xxx yyyz zzz xxx zyyyz

User · Answer

If you are only interested in the last line of input and you expect to find only one match  for example a part of the summary line of a shell command   you can also try this very compact code  adopted from How to print regexp matches using  awk       echo  xxx yyy zzz    awk   match  0  yyy  a  END print a 0    yyy   Or the more complex version with a partial result     echo  xxx a yyy b zzz c    awk   match  0  yyy          a  END print a 1    b   Warning  the awk match   function with three arguments only exists in gawk  not in mawk  Here is another nice solution using a lookbehind regex in grep instead of awk  This solution has lower requirements to your installation     echo  xxx a yyy b zzz c    grep -Po     lt  yyy         b

User · Answer

gawk can get the matching part of every line using this as action     if  match  0  your regexp  m   print m 0         match string  regexp    array     If array is present  it is cleared    and then the zeroth element of array is set to the entire portion of   string matched by regexp  If regexp contains parentheses  the   integer-indexed elements of array are set to contain the portion of   string matching the corresponding parenthesized subexpression    http   www gnu org software gawk manual gawk html String-Functions

User · Answer

This is the very basic  awk   pattern   print  0    file   ask awk to search for pattern using     then print out the line  which by default is called a record  denoted by  0  At least read up the documentation   If you only want to get print out the matched word    awk   for i 1 i lt  NF i     if  i   yyy   print  i       file

User · Answer

It sounds like you are trying to emulate GNU s grep -o behaviour  This will do that providing you only want the first match on each line   awk  match  0   regex         print substr  0  RSTART  RLENGTH      file   Here s an example  using GNU s awk implementation  gawk    awk  match  0   a t         print substr  0  RSTART  RLENGTH       usr share dict words   head act act act act aft ant apt art art art   Read about match  substr  RSTART and RLENGTH in the awk manual   After that you may wish to extend this to deal with multiple matches on the same line

User · Answer

Off topic  this can be done using the grep also  just posting it here in case if anyone is looking for grep solution  echo  xxx yyy zzze     grep -oE  yyy

User · Answer

Using sed can also be elegant in this situation  Example  replace line with matched group  yyy  from line      cat testfile xxx yyy zzz yyy xxx zzz   cat testfile   sed -r  s     yyy      1 g  yyy yyy   Relevant manual page  https   www gnu org software sed manual sed html Back 002dreferences-and-Subexpressions

[regex] How to print matched regex pattern using awk?

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