Symfony上的security.yml問題

Question

我正在嘗試使用sfGuardPlugin 5.0.0在Symfony 1.4上開發的應用程序中配置一些安全性。

要求說，管理員用戶可以做所有事情，而查看者只能列出和查看記錄。

這是我在模塊的目錄配置中使用的security.yml：

index:
    is_secure: true
    credentials: [[ admin, viewer ]]

show:
    is_secure: true
    credentials: [[ admin, viewer ]]

all:
    is_secure: true
    credentials: [ admin ]

但是我不知道為什么當我嘗試進行允許查看器執行的操作之一時，它以消息“不允許”停止。

這是堆棧跟蹤：

1    sfPatternRouting   Connect sfRoute "sf_guard_signin" (/guard/login)
2    sfPatternRouting   Connect sfRoute "sf_guard_signout" (/guard/logout)
3    sfPatternRouting   Match route "homepage" (/) for / with parameters 
  array ( 'module' => 'strain', 'action' => 'index',)
4    sfFilterChain  Executing filter "sfRenderingFilter"
5    sfFilterChain  Executing filter "sfBasicSecurityFilter"
6    Doctrine_Connection_Mysql  exec : SET NAMES 'UTF8' - ()
7    Doctrine_Connection_Statement  execute : SELECT s.id AS s__id, s.first_name AS 
 s__first_name, s.last_name AS s__last_name, s.email_address AS s__email_address, s.username 
 AS s__username, s.algorithm AS s__algorithm, s.salt AS s__salt, s.password AS s__password, 
 s.is_active AS s__is_active, s.is_super_admin AS s__is_super_admin, s.last_login AS 
 s__last_login, s.avatar AS s__avatar, s.token AS s__token, s.notify_new_order AS 
 s__notify_new_order, s.notify_ready_order AS s__notify_ready_order, s.initials AS 
 s__initials, s.created_at AS s__created_at, s.updated_at AS s__updated_at FROM sf_guard_user 
 s WHERE (s.id = ?) LIMIT 1 - (25)
8    sfBasicSecurityFilter  Action "strain/index" requires credentials 
  "[admin, viewer]", forwarding to "sfGuardAuth/secure"
9    sfFilterChain  Executing filter "sfRenderingFilter"
10   sfFilterChain  Executing filter "InboxFilter"
11   Doctrine_Connection_Statement  execute : DELETE FROM notification WHERE 
   (status = ? AND updated_at < ?) - (2, 2012-05-19 14:21:05)
12   sfFilterChain  Executing filter "sfExecutionFilter"
13   sfGuardAuthActions Call "sfGuardAuthActions->executeSecure()"
14   sfPHPView  Render "sf_app_dir/modules/sfGuardAuth/templates/secureSuccess.php"
15   main   Call "sfGuardAuth->executeSignin_form()"
16   sfPartialView  Render "sf_app_dir/modules/sfGuardAuth/templates/_signin_form.php"
17   main   Set slot "error_message"
18   sfPHPView  Decorate content with "sf_app_dir/templates/login.php"
19   sfPHPView  Render "sf_app_dir/templates/login.php"
20   main   Get slot "error_message"
21   sfWebResponse  Send status "HTTP/1.1 403 Forbidden"
22   sfWebResponse  Send header "Content-Type: text/html; charset=utf-8"

有什么線索嗎？

Answer 1

最終，我找到了問題，這不是應用程序在請求兩個憑據，而是Symfony正在查看表權限，而我正在使用該組。

非常感謝您的所有回答。

Answer 2

據堆棧跟蹤，用戶需要有兩個 admin和viewer的憑據。

如果不是這種情況，則錯誤應該是（請注意[[ ＆ ]] ）：

操作“應變/索引”需要憑據“ [[管理員，查看者]]”，轉發至“ sfGuardAuth /安全”

重新檢查您的security.yml或將完整文件粘貼到您的問題中。