bash - print regex captured groups

Question

I have a file.xml so composed:

...some xml text here...
    <Version>1.0.13-alpha</Version>
...some xml text here...

I need to extract the following information:

• mayor_and_minor_release_number --> 1.0
• patch_number --> 13
• suffix --> -alpha

I've thought the cleanest way to achieve that is by mean of a regex with grep command:

<Version>(\d+\.\d+)\.(\d+)([\w-]+)?<\/Version>

I've checked with regex101 the correctness of this regex and actually it seems to properly capture the 3 fields I'm looking for. But here comes the problem, since I have no idea how to print those fields.

cat file.xml | grep "<Version>(\d+\.\d+)\.(\d+)([\w-]+)?<\/Version>" -oP

This command prints the entire line so it's quite useless.

Several posts on this site have been written about this topic, so I've also tried to use the bash native regex support, with poor results:

regex="<Version>(\d+\.\d+)\.(\d+)([\w-]+)?<\/Version>"
txt=$(cat file.xml)
[[ "$txt" =~ $regex ]]     --> it fails!
echo "${BASH_REMATCH[*]}"

I'm sorry but I cannot figure out how to overtake this issue. The desired output should be:

1.0
13
-alpha

Answer 1

You may use this read + sed solution with similar regex as your's:

read -r major minor suffix < <(
sed -nE 's~.*<Version>([0-9]+\.[0-9]+)\.([0-9]+)(-[^<]*)</Version>.*~\1 \2 \3~p' file.xml
)

Check variable contents:

declare -p major minor suffix

declare -- major="1.0"
declare -- minor="13"
declare -- suffix="-alpha"

Few points:

• You cannot use \d without using -P (perl) mode in grep
• grep command doesn't return capture groups

Answer 2

Use this Perl one-liner:

perl -lne 'print for m{<Version>(\d+\.\d+)\.(\d+)([\w-]+)?<\/Version>};' file.xml

Example:

echo '<Version>1.0.13-alpha</Version>' | perl -lne 'print for m{<Version>(\d+\.\d+)\.(\d+)([\w-]+)?<\/Version>};'

Output:

1.0
13
-alpha

The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-n : Loop over the input one line at a time, assigning it to $_ by default.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.

SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches