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

Question

I see lots of examples and man pages on how to do things like search-and-replace using sed, awk, or gawk.

But in my case, I have a regular expression that I want to run against a text file to extract a specific value. I don't want to do search-and-replace. This is being called from bash. Let's use an example:

Example regular expression:

.*abc([0-9]+)xyz.* 

Example input file:

a b c abc12345xyz a b c 

As simple as this sounds, I cannot figure out how to call sed/awk/gawk correctly. What I was hoping to do, is from within my bash script have:

myvalue=$( sed <...something...> input.txt ) 

Things I've tried include:

sed -e 's/.*([0-9]).*/\\1/g' example.txt # extracts the entire input file sed -n 's/.*([0-9]).*/\\1/g' example.txt # extracts nothing 

Answer 1

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 

Answer 2

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'