文件名正则表达式提取

Question

I'd like to get parts of the filename

filenameblabla_2009-001_name_surname-name_surname

I'd like to get: 2009-001, name_surname, name_surname

I've come up with this, but it's no good

preg_match("/(?:[0-9-]{8})_(?:[a-z_]+)-(?:[a-z_]+)/", $filename, $matches);

Any pointers?

Thanks! BR

Answer 1

Assuming filename format doesn't change:

preg_match('#(\d{4}-\d{3})_(.*?)-(.*?)$#', $filename, $match);

Updated version to handle extension:

preg_match('#(\d{4}-\d{3})_(.*?)-(.*?)\.(.*?)$#', $filename, $match);