PHP Preg_match_all regexp pattern

Question

I need your help for a pattern i can't do myself.

I got this file :

    description1="SIOUH : Authentification Cerbere - Acces"
    description2="Acces formulaire authentification SIOUH"
    addheader="Pragma: no-cache "
    method="get"
    url="${AUTH_URL}/login/LoginDispatchAction.do?appId=${APP_ID}&backURL=${APP_URL_ENC}"
    errormessage="Phase 1 : Impossible atteindre le formulaire authentification CERBERE pour SIOUH"
    serveur="yes"
    verifypositive="loginConnectAppliForm"
    logrequest="yes" logresponse="no" />

I have this preg_match_all which works perfectly :

preg_match_all  ( '#(.*?)="(.*?)"#s', $contents, $matches, PREG_SET_ORDER ); //Chaque option

My problem is i can have simple quotes instead of double, how can i change my pattern to handle it? Example, i need to catch this line like all others:

parseresponse='name="conversationId" value="|"'

Thank a lot for your help.

Barmar · Accepted Answer · 2012-09-03 07:42:22Z

1

Use a backreference to require the closing quote to match the opening one:

'#(.*?)=([\'"])(.*?)\2#'

answered Sep 3, 2012 at 7:42

Barmar

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1 Comment

timmalos Over a year ago

Perfect ! Didn't know it was possible to do this. I was going on '#(.*?)=(?:(?:\'(.*?)\')|(?:"(.*?)"))#' but yours is better. Thank !

BSen · Accepted Answer · 2012-09-03 07:03:54Z

1

Try to use regex liek this

^(.*?)=[\'"](.*?)[\'"]$

answered Sep 3, 2012 at 7:03

BSen

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3 Comments

timmalos Over a year ago

As i can have [logrequest = "yes" logresponse = "yes"] on 1 line, i can't use ^ and $ i nmy regexp. If i use [\'"] instead of my ", i got this : Array ( [0] => parseresponse='name=" [1] => parseresponse [2] => name= ) Because i got a double quote in my simple quote (parsereponse is a pattern) which is catched and shouldn't be.

BSen Over a year ago

Then try to use greedy capture - remove ? from second group of round brackets

timmalos Over a year ago

With '#(\w*?)=["\'](.*)["\']#' i got a trouble with the [logrequest = "yes" logresponse = "yes"] which is on the same line

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