PHP Preg_match_all正则表达式模式

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.

Answer 1

尝试使用正则表达式

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

Answer 2

使用反向引用要求结束报价与开头报价相匹配：

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