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apache 中是否有任何方法可以根据您获得的 http 代码响应设置重写条件？将（302）重定向到我的身份验证提供程序的 oauth 代理（apache）前面的服务器；但是我不希望它代理 websocket 目录中的任何内容——我宁愿它 403 代替。这一切都是为了防止它不断地尝试重新验证它没有被授权并为 OpenIDC 建立大量的状态 cookie。

感谢您的考虑。

像这样的东西：

RewriteEngine On
RewriteCond %{HTTP_HOST} ^my\.server\.co$
RewriteCond %{HTTP_RESPONSE} 302
RewriteCond %{REQUEST_URI}  ^/websocket
RewriteRule (.*) $1 [F,L,NC]
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2 回答 2

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您应该能够使用：

<Location /websocket>
    OIDCUnAuthAction 401
</Location>

如配置原语中所述：

# (Optional)
# Defines the action to be taken when an unauthenticated request is made.
# "auth" means that the user is redirected to the OpenID Connect Provider or Discovery page.
# "401" means that HTTP 401 Unauthorized is returned.
# "pass" means that an unauthenticated request will pass but claims will still be passed when a user happens to be authenticated already
# Useful in Location/Directory/Proxy path contexts that serve AJAX/Javascript calls and for "anonymous access"
# When not defined the default "auth" is used.
#OIDCUnAuthAction [auth|pass|401]

（它会返回 401 状态码而不是 403）

于 2015-12-17T21:07:42.413 回答
1

其他解决方案

RewriteEngine on

ErrorDocument 403 /%{REQUEST_URI}/403.shtml
ErrorDocument 404 /%{REQUEST_URI}/404.shtml

RewriteCond %{REQUEST_URI} /([0-9]{3}+).shtml$ [NC]
RewriteRule (.*) $1 [R=%1,L]
于 2016-06-14T08:46:28.063 回答