问题
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?
回答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.
回答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.
回答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
来源：https://stackoverflow.com/questions/623861/how-do-you-unescape-urls-in-java