how to urldecode a request_uri string in Lua

问题:

When I use ngx.var.request_uri I'm getting back a string that contains %20 in place of spaces. Is there a urldecode() function or similar to decode my string?

回答1:

The decoded URI can be found in ngx.var.uri. It does not contain the query string, if you need it see ngx.var.query_string.

EDIT: if you cannot use this, here is a simple way to unescape a URL in Lua.

local hex_to_char = function(x)
  return string.char(tonumber(x, 16))
end

local unescape = function(url)
  return url:gsub("%%(%x%x)", hex_to_char)
end

Example usage:

local url = "/test/some%20string?foo=bar"
print(unescape(url)) -- /test/some string?foo=bar

But you should probably split the query string before using it.

回答2:

If you are using nginx-lua-module then you can use below api for this.

newstr = ngx.unescape_uri(str)

You can also take a look of ngxescape_uri