I have some data that looks like the following.
\\0101 \\0102 \\0103 \\0104
\\0201 \\0202 \\0203 \\0204 \\0205 \\0206
\\0301 \\0302 \\0303 \\0304 \\0305 \\0306
I need to always get the digits before the last \\ in the line.
So in the above lines, my output should be.
0103
0205
0305
I am matching these digits, but its also matching the last set.
(?<=\\\\)\d+(?: \\\\\d+$)
How can I exclude everything else except those digits?
You were close, instead of using a non-capturing group (?: use a Positive Lookahead.
(?<=\\\\)\d+(?= +\\\\\d+$)
See live demo
There's a many ways to do that, but to correct your regex, just add a capturing group on the target digits
(?<=\\\\)(\d+)(?: \\\\\d+$)
And another one (wich could be simplified more):
\\\\(\d+)\s+[\d\\]+$
If your data are always in the same form, the lookbehind is useless. You can try this:
$subject = <<<'LOD'
\\0101 \\0102 \\0103 \\0104
\\0201 \\0202 \\0203 \\0204 \\0205 \\0206
\\0301 \\0302 \\0303 \\0304 \\0305 \\0306
LOD;
preg_match_all('~\d+(?=\D+\d+$)~m', $subject, $matches);
print_r($matches);