How do you unescape URLs in Java?

Question

When I read the xml through a URL's InputStream, and then cut out everything except the url, I get "http://cliveg.bu.edu/people/sganguly/player/%20Rang%20De%20Basanti%20-%20Tu%20Bin%20Bataye.mp3".

As you can see, there are a lot of "%20"s.

I want the url to be unescaped.

Is there any way to do this in Java, without using a third-party library?

Answer 1

This is not unescaped XML, this is URL encoded text. Looks to me like you want to use the following on the URL strings.

URLDecoder.decode(url);

This will give you the correct text. The result of decoding the like you provided is this.

http://cliveg.bu.edu/people/sganguly/player/ Rang De Basanti - Tu Bin Bataye.mp3

The %20 is an escaped space character. To get the above I used the URLDecoder object.

Answer 2

URLDecoder.decode(String s) has been deprecated since Java 5

You should use URLDecoder.decode(String s, String enc).

For example:

URLDecoder.decode(url, "UTF-8")

Regarding the encoding to use:

Note: The World Wide Web Consortium Recommendation states that UTF-8 should be used. Not doing so may introduce incompatibilites.

Answer 3

I'm having problems using this method when I have special characters like á, é, í, etc. My (probably wild) guess is widechars are not being encoded properly... well, at least I was expecting to see sequences like %uC2BF instead of %C2%BF.

Edited: My bad, this post explains the difference between URL encoding and JavaScript's escape sequences: URI encoding in UNICODE for apache httpclient 4