URISyntaxException - How to deal with urls with %

Question

I am fairly new to Java and came across this issue. I tried searching but never got a correct answer.

I have a string for example

String name = anything 10%-20% 04-03-07

Now I need to build up a url string with this String name like below.

http://something.com/test/anything 10%-20% 04-03-07

I tried replacing the spaces with %20 and now I am getting the new url as

http://something.com/test/anything%2010%-20%%2004-03-07

When I use this url and fire it in firefox it just works fine but while processing in Java it is apparently throwing

Exception in thread "main" java.lang.IllegalArgumentException
at java.net.URI.create(Unknown Source)
at org.apache.http.client.methods.HttpGet.<init>(HttpGet.java:69)
Caused by: java.net.URISyntaxException: Malformed escape pair at index 39 : 
at java.net.URI$Parser.fail(Unknown Source)
at java.net.URI$Parser.scanEscape(Unknown Source)
at java.net.URI$Parser.scan(Unknown Source)
at java.net.URI$Parser.checkChars(Unknown Source)
at java.net.URI$Parser.parseHierarchical(Unknown Source)
at java.net.URI$Parser.parse(Unknown Source)
at java.net.URI.<init>(Unknown Source)
... 6 more

This is the code throwing error

HttpClient httpclient = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
HttpResponse response = httpclient.execute(httpget);

Answer 1

Encode also the percent sign with %25.

http://something.com/test/anything 10%-20% 04-03-07 would work with http://something.com/test/anything%2010%25-20%25%2004-03-07.

You should be able to use for example URLEncoder.encode for this - just remember, that you need to urlencode the path part, not anything before that, so something like

String encodedUrl =
    String.format("http://something.com/%s/%s",
      URLEncoder.encode("test", "UTF-8"),
      URLEncoder.encode("anything 10%-20% 04-03-07", "UTF-8")
    );

Note: URLEncoder encodes spaces to + instead of %20, but it should work equally well, both are ok.