how to use sed, awk, or gawk to print only what is matched? how to use sed, awk, or gawk to print only what is matched? unix unix

how to use sed, awk, or gawk to print only what is matched?


My sed (Mac OS X) didn't work with +. I tried * instead and I added p tag for printing match:

sed -n 's/^.*abc\([0-9]*\)xyz.*$/\1/p' example.txt

For matching at least one numeric character without +, I would use:

sed -n 's/^.*abc\([0-9][0-9]*\)xyz.*$/\1/p' example.txt


You can use sed to do this

 sed -rn 's/.*abc([0-9]+)xyz.*/\1/gp'
  • -n don't print the resulting line
  • -r this makes it so you don't have the escape the capture group parens().
  • \1 the capture group match
  • /g global match
  • /p print the result

I wrote a tool for myself that makes this easier

rip 'abc(\d+)xyz' '$1'


I use perl to make this easier for myself. e.g.

perl -ne 'print $1 if /.*abc([0-9]+)xyz.*/'

This runs Perl, the -n option instructs Perl to read in one line at a time from STDIN and execute the code. The -e option specifies the instruction to run.

The instruction runs a regexp on the line read, and if it matches prints out the contents of the first set of bracks ($1).

You can do this will multiple file names on the end also. e.g.

perl -ne 'print $1 if /.*abc([0-9]+)xyz.*/' example1.txt example2.txt