允许在security.yml中对Symfony对象操作进行不安全的访问(Allow unsecured access to Symfony object action in security.yml)

 我在Symfony 1.4.13中有一个特定的模块，默认设置在security.yml文件中。 我有一个特定的对象动作，我希望任何人都可以访问（注销用户），但似乎无法找到正确的方法来在YAML文件中找出动作的名称以获得匹配。 
 
 具体来说，我有一个project模块，其中包含典型的索引，显示，创建等操作，以及runReport对象操作（因此操作方法的名称为executeListRunReport ）。 security.yml文件如下： 
 
all:
  is_secure: true

index:
  credentials: pm_view

show:
  credentials: pm_view

filter:
  credentials: pm_view

runReport:   # This is the one that is giving me problems
  is_secure: false
 
 actions.php中的方法是： 
 
public function executeListRunReport(sfWebRequest $request) {
...
}
 
 当进入project/[idOfObject]/ListRunReport时，这适用于登录用户。 
 
 如何编写security.yml文件以允许任何人访问该操作（例如直接从我手工生成的URL）而无需登录？ 谢谢！ 

I have a particular module in Symfony 1.4.13 that is set to be secured by default in its security.yml file. I have one particular object action that I would like accessible by anyone (logged out users), but can't seem to find the correct way to right out action's name in the YAML file to get a match on it.
 
Specifically, I have a project module with the typical index, show, create, etc. actions, as well as a runReport object action (so the action method's name is executeListRunReport). The security.yml file is as follows:
 
all:
  is_secure: true

index:
  credentials: pm_view

show:
  credentials: pm_view

filter:
  credentials: pm_view

runReport:   # This is the one that is giving me problems
  is_secure: false
 
My method in actions.php is:
 
public function executeListRunReport(sfWebRequest $request) {
...
}
 
This works just fine for a logged in user when going to project/[idOfObject]/ListRunReport. 
 
How can I write the security.yml file to allow anyone to access that action (directly from a URL that I generate by hand for example) without having to log in? Thanks!

原文：https://stackoverflow.com/questions/7810210