Play Framework双网址编码

Question

given the following controller method where username = bob and emailAddress = bob@bob.com

    public static void resetPassword(String username, String emailAddress) {

            String url = BASE_URL + "/users/" + username + "/reset_password";

            HttpResponse response = WS.url(url).setParameter("email_address", emailAddress).get();
}

Sometimes when I make the call the url endpoing receives:

localhost:8080/api/v1/users/bob/reset_password?email_address= bob%40bob.com

then other times i get: localhost:8080/api/v1/users/bob/reset_password?email_address= bob%2540bob.com

On the second one the @ has been encoded once to %40 then the % was again encoded to %25 so you end up with %2540

If I do nothing more than wait a minute the problem goes away which makes me think it's some sort of caching problem but I can't seem to figure out what it is.

Answer 1

最终被认为是一个错误，并已在以后的版本中修复

Answer 2

since this does not repeduce all the times i dont know if that will help but did you try to encode the url? for example : org.apache.commons.httpclient.util.URIUtil.encodeAll(url);

Answer 3

I think what is happening is when you are calling the controller the first time you send the at symbol as an at symbol.

This then gets encoded into the response and the @ is converted to %40. I'm guessing when you get this back and you resend it through the browser the % in the %40 gets encoded to %25 making it %2540.

Sorry if the above is confusing, it is difficult to explain.

Simple answer would be to just do a replace on the emailAddress variable of %40 to @ before passing it into WS class. There is also a urlDecode() method in the play framework which may do the trick, I've used play before but I haven't used this method.

Answer 4

I suspect some form of URL rewriting. Do you have an Apache Web Server running in front of your server? Maybe some RewriteRule is missing the [NE] flag.