I have the following regex:
$string = 'font-size:12em;';
$pattern = '@:[A-Za-z0-9.]+;@i';
preg_match($pattern, $string, $matches);
$matches returns:
Array ( [0] => :12em; )
However, why is the : and ; returned? How can I get it to not return those colons and only return the CSS value 12em?
Because the first element in that array is the whole match. Use a capturing group, and the second element (or use lookarounds).
Example:
preg_match('/:\s*(\w[^;}]*?)\s*[;}]/', $string, $matches);
print $matches[1];
Note that things like these will not work in all cases. Comments and more complicated statements could break it.
Example:
/* foo: bar; */
foo: url("bar?q=:x;");
Use this pattern instead:
@(?<=:)[A-Za-z0-9.]+(?=;)@i
The explanation is that you the (?<=) and (?=) are respectively lookbehind and lookahead groups. Which means they aren't captured as part of your match.
Edit For handling %'s +more
@(?<=:)[^;]+@i
width:10% it doesn't seem to return any value? Maybe the % is a special character?