checking version in shell script using regex

Question

Could anybody suggest a reg ex for checking versions What if I like to check any update version in 1.0.0 release may be 1.0.0-1 or 1.0.0-2 or 1.0.0-3 and I just need to check for what update version is it -1 or -2 or -3 1.0.0-1 or 1.0.0-2 or 1.0.0-3 what regex can I use ? tried

sed -ne s/1.0.0\-[0-9]\1/p

Thanks !

Answer 1

Perhaps something like this?:

$ echo "1.0.0-1" | sed -ne 's/1.0.0\(\-[0-9]\)/\1/p'
-1

The regex to get the update number for any version would be:

$ echo "1.0.0-1" | sed -ne 's/[0-9]\+\.[0-9]\+\.[0-9]\+\(\-[0-9]\)/\1/p'
-1

The trick is to capture the output you want between \( and \) grouping parentheses and the use \1 to output the pattern in the replace part of the command.

Answer 2

You don't really need a regular expression here; just split the string on the correct character:

version=1.0.0-2
major=$(echo $version | cut -d. -f1 )
minor=$(echo $version | cut -d. -f1 )
patchlvl=$(echo $version | cut -d. -f1 | cut -d - -f1 )
build=$(echo $version | cut -d - -f2)

Answer 3

This will leave only the update version number:

echo "1.0.0-1" | sed -e 's/^[1-9][0-9.]*-//'

And this will leave the leading '-' on it:

echo "1.0.0-1" | sed -e 's/^[1-9][0-9.]*//'

The p command isn't needed in the above so I left it, and the -n option off, but you can restore them if this part of some more elaborate sed script.

Edit: OP apparently just wants to know if the version contains an update. Here are a couple of ways in classic bourne shell:

#!/bin/sh
version='1.0.0-1'
case "$version" in
    *-[1-9]*)  # Not a regex, this is a (file) "pattern"
        # Update triggered code here
        ;;
esac

# Or:
if echo "$version" | grep '-[1-9][0-9]*$' >/dev/null ; then
    # Update triggered code here
fi

There are other ways to do it in other shells/environments of course.