我需要使用优秀的库https://github.com/openresty/lua-nginx-module
将Nginx变量传递给我的PHP 7.0后端我更喜欢使用content_by_lua_block
而不是set_by_lua_block
,因为'set'函数的文档声明“此指令旨在执行短的,快速运行的代码块,因为Nginx事件循环被阻止因此,在代码执行期间,应避免使用耗时的代码序列。“
https://github.com/openresty/lua-nginx-module#set_by_lua
但是,由于'content _...'函数是非阻塞的,因此下面的代码不会及时返回,并且传递给PHP时会取消设置$ hello:
location ~ \.php{
set $hello '';
content_by_lua_block {
ngx.var.hello = "hello there.";
}
fastcgi_param HELLO $hello;
include fastcgi_params;
...
fastcgi_pass unix:/run/php/php7.0-fpm.sock;
}
问题是,如果采用某些代码路径,我的Lua代码有可能成为“耗时的代码序列”,例如使用加密。
以下Nginx位置工作得很好,但那是因为set_by_lua_block()是一个阻塞函数调用:
location ~ \.php {
set $hello '';
set_by_lua_block $hello {
return "hello there.";
}
fastcgi_param HELLO $hello;
include fastcgi_params;
...
fastcgi_pass unix:/run/php/php7.0-fpm.sock;
}
我的问题是,这里最好的方法是什么?有没有办法在我的变量设置后才能从content_by_lua_block()中调用Nginx指令fastcgi_pass
和相关指令?
答案 0 :(得分:1)
是的,可以使用ngx.location.capture
。写一个单独的位置块,例如:
location /lua-subrequest-fastcgi {
internal; # this location block can only be seen by Nginx subrequests
# Need to transform the %2F back into '/'. Do this with set_unescape_uri()
# Nginx appends '$arg_' to arguments passed to another location block.
set_unescape_uri $r_uri $arg_r_uri;
set_unescape_uri $r_hello $arg_hello;
fastcgi_param HELLO $r_hello;
try_files $r_uri =404;
fastcgi_split_path_info ^(.+\.php)(/.+)$;
include fastcgi_params;
fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
fastcgi_param SCRIPT_NAME $fastcgi_script_name;
fastcgi_index index.php;
fastcgi_pass unix:/run/php/php7.0-fpm.sock;
}
然后你可以这样打电话:
location ~ \.php {
set $hello '';
content_by_lua_block {
ngx.var.hello = "hello, friend."
-- Send the URI from here (index.php) through the args list to the subrequest location.
-- Pass it from here because the URI in that location will change to "/lua-subrequest-fastcgi"
local res = ngx.location.capture ("/lua-subrequest-fastcgi", {args = {hello = ngx.var.hello, r_uri = ngx.var.uri}})
if res.status == ngx.HTTP_OK then
ngx.say(res.body)
else
ngx.say(res.status)
end
}
}