将此Apache重写转换为Nginx规则

Question

I need to convert this .htaccess rules to nginx rules:

RewriteEngine on
Options +FollowSymlinks
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ navigation.php
RewriteRule .* - [E=HTTP_X_REQUESTED_WITH:%{HTTP:X-Requested-With}]

I used this online conversion tool, which gives:

if (!-f $request_filename){
    set $rule_0 1$rule_0;
}
if (!-d $request_filename){
    set $rule_0 2$rule_0;
}
if ($rule_0 = "21"){
    rewrite ^/(.*)$ /navigation.php;
}
     HTTP_X_REQUESTED_WITH:$http_x_requested_with;
#ignored: "-" thing used or unknown variable in regex/rew 

However the result is not usable in nginx 0.76, and I got an error saying

unknown directive "setenv" 

And when I eliminate the last line from my nginx config, the script fails to show some pages.

So I really appreciate if you could do the conversion.

Answer 1

With Nginx you do not need to remap exactly the way you did it in Apache.

Did you ever read the IfIsEvil nginx page?

To test a file is not a static file or directory before applying a rule in nginx the real simple way is not a multi-if like you did but a try_file directive ( check if file exists in pitfall page ).

So you should start with that "rule":

server {
  root /var/www/domain.com;
  location / {
    HTTP_X_REQUESTED_WITH:$http_x_requested_with;
    try_files $uri $uri/ /navigation.php;
  }
}

Now you have maybe some other locations to handle PHP files via fastcgi, this is maybe the real place where you need to transfer the HTTP_X_REQUESTED_WITH as a fastcgi_param .