Question

I have a problem involving PHP and regex.

I have some files and need to capture some of the names. But there’s one part that are not in all.

Example files:
D.SAE.IND.001.TME.BR.SER.20120729
D.SUI.IND.003.TMC.GER.TOT.201206
D.SAE.LIS.008.AGE.APS.SER.45D.20120713

Note that the last file has a part (“45D”) that is optional.
I need to get every group. I was using the following regex:

preg_match('/D\.(?P<font>[\w]*)\.(?P<tipo>[\w]*)\.(?P<numberLevel>[\d]*)\.(?P<indicator>[\w]*)\.(?P<nameLevel>[\w]*)\.(?P<group>[\w]*)\.(?P<id>[\d]*)/i', $arrInput , $result);

The problem is that optional part (eg the latest file – “45D”).
I need this result:

D.SAE.IND.001.TME.BR.SER.20120729
$result[‘fonte’] = ‘SAE’
$result[‘tipo’] = ‘IND’
$result[‘numberLevel’] = ‘001’
$result[‘indicator’] = ‘TME’
$result[‘nameLevel’] = ‘BR’
$result[‘group’] = ‘SER’
$result[‘op’] = ”
$result[‘id’] = ‘20120729’

D.SAE.LIS.008.AGE.APS.SER.45D.20120713
$result[‘fonte’] = ‘SAE’
$result[‘tipo’] = ‘LIS’
$result[‘numberLevel’] = ‘008’
$result[‘indicator’] = ‘AGE’
$result[‘nameLevel’] = ‘APS’
$result[‘group’] = ‘SER’
$result[‘op’] = ’45D’
$result[‘id’] = ‘20120713’

Thank you in advance for help.

Answer 1

This works for me:

/D\.(?P<font>[\w]*)\.(?P<tipo>[\w]*)\.(?P<numberLevel>[\d]*)\.(?P<indicator>[\w]*)\.(?P<nameLevel>[\w]*)\.(?P<group>[\w]*)\.(?:(?P<op>.*)\.)?(?P<id>[\d]*)/i

The key is adding:

(?:(?P<op>.*)\.)?

Before the id group.

Edit: Added the ?: to make the grouping non-capturing.