how can i get numb开发者_如何学Pythoner ID in action? for example class jobActions extends sfActions
I would like to be able to user sfPropelPager with sfGuard\'s user ta开发者_运维知识库ble. I can do the following just fine:
I use symfony 1.4.11 with Doctrine. I have site, and it has 3 languages. I use sfDoctrineGuardPlugin 4.0.1 , and I have 3 groups of users. In site I show to each user his group, and I needit show to e
I use symfony 1.4.11/ U开发者_JAVA技巧ser on my site can add links to post like: http://stackoverflow.com/ , www.stackoverflow.com/ , stackoverflow.com/
I am running Symphony 1.4 and on the following tutorial: http://www.symfony-project.org/jobeet/1_4/Doctrine/en/03.
I\'m trying to integrate CAS and sfDoctrineGuard. I want my login and logout to be controlled by CAS, which works perfectly with the sfCASPlugin.
i have a little problem with symfony versionable feature in Doctrine one-to-many relationships. i have one-to-many relationship and i want to make it versionable. here is my schemas:
How can I check any user\'s password in Symfony? (not only the logged in user) (I use sfGuardPlugin) I\'ve tried like this, but doesn\'t work:
Does anyone have any of how to force a login action with sfGuard th开发者_运维问答rough code? Something like: $this->authenticate(\'username\'); ?
I have a little disturbing problem in here! an using symfony 1.4 with Doctrine! i fact i have a \"many to many\" relation (see code bellow) but i don\'t have the RIGHT result!
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