CakePHP在一个使用nginx的子目录中(重写规则？)

Question

我设法让这个工作回来了一段时间,但是回到我开始的cakephp项目时,似乎我最近对nginx所做的任何改变(或者最近的更新)都破坏了我的重写规则.

目前我有:

worker_processes  1;

events {
    worker_connections  1024;
}


http {
    include       mime.types;
    default_type  application/octet-stream;

    sendfile        on;

    keepalive_timeout  65;

    server {
        listen       80;
        server_name  localhost;

        location / {
            root   html;
            index  index.php index.html index.htm;
        }

        location /basic_cake/ {
            index  index.php;

            if (-f $request_filename) {
              break;
            }
            if (!-f $request_filename) {
              rewrite ^/basic_cake/(.+)$ /basic_cake/index.php?url=$1 last;
              break;
            }
        }

        location /cake_test/ {
            index  index.php;

            if (-f $request_filename) {
              break;
            }
            if (!-f $request_filename) {
              rewrite ^/cake_test/(.+)$ /cake_test/index.php?url=$1 last;
              break;
            }
        }

        # redirect server error pages to the static page /50x.html
        #
        error_page   500 502 503 504  /50x.html;
        location = /50x.html {
            root   html;
        }

        # pass the PHP scripts to FastCGI server listening on 127.0.0.1:9000
        #
        location ~ \.php$ {
            root           html;
            fastcgi_pass   127.0.0.1:9000;
            fastcgi_index  index.php;
            fastcgi_param  SCRIPT_FILENAME  $document_root$fastcgi_script_name;
            include        fastcgi_params;
        }

    }

    server {
        listen       8081;
        server_name  localhost;

        root /srv/http/html/xsp;

        location / {
            index  index.html index.htm index.aspx default.aspx;
        }

        location ~ \.(aspx|asmx|ashx|asax|ascx|soap|rem|axd|cs|config|dll)$ {
            fastcgi_pass   127.0.0.1:9001;
            fastcgi_param  SCRIPT_FILENAME  $document_root$fastcgi_script_name;
            include        fastcgi_params;
        }

    }

}

我遇到的问题是css和图像不会从webroot加载.相反,如果我访问http://localhost/basic_cake/css/cake.generic.css,我会得到一个页面告诉我:

CakePHP:快速开发php框架Missing Controller

错误:找不到CssController.

错误:在文件中创建下面的类CssController:app/controllers/css_controller.php

注意:如果要自定义此错误消息,请创建app/views/errors/missing_controller.ctp CakePHP:快速开发php框架

有没有人对如何解决这个问题有任何想法？

(我在ServerFault上发布了这个,但我有这种感觉,与此相比,没有多少人检查该网站,所以我可能不应该打扰...)

Answer 1

使用别名和try_files指令可以更轻松地完成此操作.我从使用PHP的简单配置开始,并在服务器上的path/cakeproject中添加了一个Cake项目:

    root /var/www;
    index index.php;

    location /cakeproject {
            alias /var/www/cakeproject/app/webroot;
            try_files $uri $uri/ /cakeproject/app/webroot/index.php;
    }

    location ~ \.htaccess {
            deny all;
    }

    location ~ \.php$ {
            fastcgi_pass 127.0.0.1:9000;
            fastcgi_index index.php;
            fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
            include /etc/nginx/fastcgi_params;
    }

蛋糕项目现在可以从http://thedomain.com/cakeproject/完美运行

Answer 2

我设法通过将cake.base参数添加到cakephp配置而不是使用App.baseUrl(查看dispatcher.php中的代码)来解决这个问题.

所以,假设我有/ var/www/html/cakeprj中的cakephp副本,我的WWWROOT是/ var/www/html:

1. nginx主机配置

   # that's for all other content on the web host
   location / {
       root        /var/www/html;
       autoindex   off;
       index       index.php index.html index.htm;
   
       ...
   }
   
   # that's for cakephp
   location /cakeprj {
       rewrite     ^/cakeprj$ /cakeprj/ permanent;
       rewrite     ^/cakeprj/(.+)$ /$1 break;
       root        /var/www/html/cakeprj/app/webroot;
       try_files   $uri /$uri/ @cakephp;
   }
   
   # that's for all other php scripts on the web host
   location ~ \.php$ {
       root                                /var/www/html;
       fastcgi_pass                        unix:/var/lib/fcgi/php-fcgi.socket;
       ...
       include                             /etc/nginx/fastcgi_params;
   }
   
   
   # that's for cakephp execution
   location @cakephp {
       set $q $request_uri;
       if ($request_uri ~ "^/cakeprj(.+)$") {
           set $q $1;
       }
       fastcgi_param SCRIPT_FILENAME       /var/www/html/cakeprj/app/webroot/index.php;
       fastcgi_param QUERY_STRING          url=$q;
       fastcgi_pass                        unix:/var/lib/fcgi/php-fcgi.socket;
       include                             /etc/nginx/fastcgi_params;
   }
   

2. app/config/core.php中的cakephp配置

   Configure::write('App.base', '/cakeprj');
   Configure::write('App.baseUrl', '/cakeprj/'); // seems like it doesn't matter anymore
   

...并且瞧 - 你正确地为nginx服务cakephp静态文件,正确地传递给cakephp调度程序的url以及正确生成url的cakephp.

PS如果你的nginx不支持try_files我相信它的配置可以用if条件和另一个重写来重写.