Q:

Characters problem in Bit.ly

When I try to shorten a link with "#,&" character I get an exception.  Is there a way to handle these character properly? 
This is a sample code that works:
String shortUrl = bitly.getShortUrl("http://z"); //Works

If I add for example '&' or '%25' to the string it will throw an exception:
String shortUrl = bitly.getShortUrl("http://z%26"); // Exception 
String shortUrl = bitly.getShortUrl("http://z&"); // Exception

The getShortUrl function from this Java class. 
Thanks

A:

That library (the Java class you link to) doesn't escape the URL... that's pretty awful. 
Excerpt:
private String getBitlyHttpResponseText(String urlToShorten) throws IOException {
  String uri = getBitlyUrl() + urlToShorten + bitlyAuth;
  HttpGet httpGet = new HttpGet(uri);
  ...

Notice how urlToShorten isn't escaped in any way, shape or form.  Prone to injection-style attacks, and just generally doesn't work. 
Anyway, you'll need to escape urlToShorten.