Java URL param replace %20 with space

Question

In my web page when form is submitted witha space entered in text field, it is being read as %20 in backend java code instead of space. I can replace %20 back to "" in backend but i think it is not the right approach and it could happen anywhere in the application.

Is there any better way of handling it in front end when you submit form?

Answer 1

That's nothing wrong with that. It's how characters are escaped in a URL. You should use URLDecoder, which is particularly appropriate because, despite of its name, it does application/x-www-form-urlencoded decoding:

String decoded = URLDecoder.decode(queryString, "UTF-8");

Then you'll be able to build a map of key/value pairs parsing the query part of the URL, splitting on &, and using = to separate the key from the value (which may also be null).

However note that if the URL is passed as a URI object, it has a nice getQuery() which already returns the unescaped text.

If you use the servlet API, you don't have to escape anything because there are nice methods like getParameterMap().

Answer 2

You could pass it through a the URLDecoder, that way your are not just sorting the problem for %20 but other URLEncoded values http://docs.oracle.com/javase/7/docs/api/java/net/URLDecoder.html