Nginx 重写目录并排除子目录

Question

This is my current nginx config file.

server {
    listen 6063;

    root /var/www/html/project;
    index index.php;

    location ~ \.php$ {
        include snippets/fastcgi-php.conf;
        fastcgi_pass unix:/run/php/php7.2-fpm-backend.sock;
    }    

    rewrite ^/admin/(.*)$ /admin/index.php break;
}

Any access to /admin will rewrite to /admin/index.php.

Now for some reasons I have to add css folder, js folder to /var/www/html/project/admin/includes.

So the new requires is any access to /admin/includes/* will access the real files for css and js. All other access /admin/* (not /admin/includes) will rewrite to /admin/index.php.

Or I can say any access to folder /admin if doesn't exists (like /admin/includes/css/example.css), then rewrite to /admin/index.php

Thanks,

Answer 1

OK, I found the answer.

location /admin {
    if (!-e $request_filename) {
        rewrite (.*)$ /admin/index.php last;
    }
}

If any access to folder /admin which doesn't exists ( it is not a real file like css or js files) will rewrite to /admin/index.php.

Answer 2

Nginx uses the longest matching location ( see docs ), so it's enough to give a more specific location directive:

server {
    …

    location ~ /admin/css {
    }

    location ~ /admin {
       rewrite (.*)$ /admin/index.php last;
    }
}

Then, if anything matches /admin/css , it will process only the empty block (and so get served by default), and all that matches /admin (and not /admin/css ) will hit the rewrite.