Finding multiple strings on multiple lines in file and manipulating output with bash/perl - perl

Question

I am trying to get the version numbers for content management systems being hosted on my server. I can do this fairly simply if the version number is stored on one line with something like this:
grep -r "\$wp_version = '" /home/
Which returns exactly what I want to stdout:
/home/$RANDOMDOMAIN/wp-includes/version.php:$wp_version = '3.7.1';
The issue I run into is when I start looking for version numbers that are stored on two or more lines, like Joomla! or Magento which use the following formats respectively:
Joomla!:
/** #var string Release version. */
public $RELEASE = '3.2';
/** #var string Maintenance version. */
public $DEV_LEVEL = '3';
Magento:
'major' => '1',
'minor' => '8',
'revision' => '1',
'patch' => '0',
I have gotten it to 'work', in a way, using the following (With this method if, for whatever reason, one of the strings I am looking for is missing the whole command becomes useless since xargs -l3 is expecting 2 rows above the path provided by -print):
find /home/ -type f -name version.php -exec grep " \$RELEASE " '{}' \; -exec grep " \$DEV_LEVEL " '{}' \; -print | xargs -l3 | sed 's/\<var\>\s//g;s/\<public\>\s//g' | awk -F\; '{print $3":"$1""$2}' | sed 's/ $DEV_LEVEL = /./g'
Which get's me output like this:
/home/$RANDOMDOMAIN/version.php:$RELEASE = 3.2.3
/home/$RANDOMDOMAIN/anotherfolder/version.php:$RELEASE = 1.5.0
I also have a working for loop that WILL exclude any file that does not contain both strings, but depending how much it has to sift through, can take significantly longer than the find one liner above:
for path in $(grep -rl " \$RELEASE " /home/ 2> /dev/null | xargs grep -rl " \$DEV_LEVEL ")
do
joomlaver="$path"
joomlaver+=$(grep " \$RELEASE " $path)
joomlaver+=$(echo " \$DEV_LEVEL = '$(grep " \$DEV_LEVEL " $path | cut -d\' -f2)';")
echo "$joomlaver" | sed 's/\<var\>\s//g;s/\<public\>\s//g;s/;//g' | awk -F\' '{ print $1""$2"."$4 }' | sed 's/\s\+//g'
unset joomlaver
done
Which get's me output like this:
/home/$RANDOMDOMAIN/version.php$RELEASE=3.2.3
/home/$RANDOMDOMAIN/anotherfolder/version.php$RELEASE=1.5.0
But I have to believe there is a simpler, shorter, more elegant way. Bash is preferred or if it can somehow be done with a perl one liner, that would work as well. Any and all help would be much appreciated. Thanks in advance. (Sorry for all the edits, but I am trying to figure this out myself as well!)

Answer 1

Here is a perl one-liner that will extract the $RELEASE and $DEV_LEVEL from the php file format you showed:
perl -ne '$v=$1 if /\$RELEASE\s*=\s*\047([0-9.]+)\047/; $devlevel=$1 if /\$DEV_LEVEL\s*=\s*\047([0-9.]+)\047/; if (defined $v && defined $devlevel) { print "$ARGV: Release=$v Devlevel=$devlevel\n"; last; }'
The -n makes perl effectivly wrap the whole thing inside a while (<>) { } loop. Each line is checked against two regexes. If both of them have matched then it will print the result and exit.
The \047 is used to match single quotes, otherwise the shell would get confused.
If it does not find a match, it does not print anything. Otherwise it prints something like this:
sample.php: Release=3.2 Devlevel=3
You would use it in combination with find and xargs to traverse down a directory structure, perhaps like this:
find . -name "*.php" | xargs perl -ne '$v=$1 if /\$RELEASE\s*=\s*\047([0-9.]+)\047/; $devlevel=$1 if /\$DEV_LEVEL\s*=\s*\047([0-9.]+)\047/; if (defined $v && defined $devlevel) { print "$ARGV: Release=$v Devlevel=$devlevel\n"; last; }'
You could make a similar version for the other file format (Magento?) you mentioned.