I need to select and output into a file some text contained in a specific string.

Let's say the string is: ABCDEFGHIJKLMNOP

What would be a command to extract what is after ABCDEF and before K? (i.e. GHIJ only) to another file?

I tried with grep command but due to my poor understanding of its complexity it failed every time. I must be missing something very basic.

Thanks a lot in advance.


I am assuming that you are programming using a shell, such as bash.

You will need sed.


sed -re 's/^ABCDEF(.*)K.*/\1/'

This uses grep to search and replace s/thing to find as regexp/thing to replace it with/

\1 means replace with 1st bracketed expression.

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  • 1
    You don't need -e when you're using -r. And -r is GNU extension. There isn't GNU sed everywhere. -r is easier for understanding, but without it script is more portable. So, it can be sed 's/^ABCDEF\(.*\)K.*/\1/'. – ValeriyKr Jan 7 '17 at 14:42

On finding a match, grep prints the whole matched line. To make it print only the matched text, you can use -o switch. But, this prints the whole matched string within the line. To make it print what you need, you should use look-behind and look-ahead assertions. This requires a grep with PCRE support.

> grep -Po '(?<=ABCDEF).+(?=K)' <<< 'ABCDEFGHIJKL'


  • -P Enable perl compatible regular expression
  • -o Print only the matched text, not whole line
  • (?<=ABCDEF) Positive look-behind assertion for matching ABCDEF
  • (?=K) Positive look-ahead assertion for matching K
  • .+ Match everything between the above assertions
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There is nothing wrong with either of the solutions posted, but you can do this in pure Bash if you like.

Given a string with your value:

echo $myStr 

You can do:

echo $(expr match "$myStr" '.*F\(.*\)K.*')
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