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How do I perform a regex match on a file and output the result.

For example, find version="X.Y.Z" and store in a variable the matched result of X.Y.Z so as to be later able to echo the match.

The is as far as I've gotten on my own: sed -n -e 's/version\=\".*\"' $PATH"version.js"

My file contents are:

var config;
function check() {
    if (window.environment && window.revision) {
        config.version = "0.0.1";
        config.isApp = true;
    }
    else {
        setTimeout(check);
    }
}
check();

and I wish to store 0.0.1 in a variable as I wish to insert this value into another file.

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  • Can you clarify what output you are piping (i assume) into sed ? Are the files you are searching in one folder ? is the version in the file name, or in the file itself ? Commented Dec 8, 2016 at 8:53
  • @kyze I have just one file which I wish to pipe in: $PATH"version.js" Commented Dec 8, 2016 at 8:58
  • Alright, so you are outputting the contents of the file and are regex-searching the contents ? Would a cat version.js | grep ... not do the trick in this case ? Commented Dec 8, 2016 at 9:03
  • @kyze I'm very new to bash so I will explore your suggestion, thanks! Commented Dec 8, 2016 at 9:06
  • If you give some example contents of the file, i (and others too) can assist. Commented Dec 8, 2016 at 9:09

1 Answer 1

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This should be working in your case:

grep config.version /path/to/version.js | grep -P -o '\d\.\d\.\d'

If your number is ever gonna be double-digit (or more) you need to add a + to the \d corresponding to the number.

As '\d\.\d\.\d' matches 1.1.1, 1.1.2, 2.2.9 etc. it would not match 1.10.2, then the regex would need to be modified to '\d\.\d+\.\d'.

You could go the "easy/safe" route and just put '\d+\.\d+\.\d+'.

The output of the first would just be your version number, which could be set to a variable after.

As an edit to your comment: This works correctly for me. I have created a file with your given inputs, grepping this with my command yields this:

enter image description here

Using this to set a variable as such, then echoing works as well:

enter image description here

enter image description here

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  • Thanks for your help @kyze. I tried: temp=$(cat $BUILD_PATH"config.js" | grep config.version | grep -P -o '\d\.\d\.\d') echo $temp but it didn't work. I also tried temp=$(cat $BUILD_PATH"config.js" | grep config.version) and it outputs config.version = "0.0.1"; Commented Dec 8, 2016 at 9:31
  • My grep is as follows: usage: grep [-abcDEFGHhIiJLlmnOoqRSsUVvwxZ] [-A num] [-B num] [-C[num]] [-e pattern] [-f file] [--binary-files=value] [--color=when] [--context[=num]] [--directories=action] [--label] [--line-buffered] [--null] [pattern] [file ...] Commented Dec 8, 2016 at 9:34
  • Ah it works if I remove the P flag and use asterisk instead of plus. Commented Dec 8, 2016 at 9:40
  • It works if you remove the -P ? This should not be happening, as standard grep can not interpret the \d regex and would need [[:digit:]] Commented Dec 8, 2016 at 9:43
  • Ok, weird. I'm using GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin16). My grep is now $(cat $BUILD_PATH"config.js" | grep config.version | grep -o '\d*\.\d*\.\d*') Commented Dec 8, 2016 at 9:46

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