views:

61

answers:

3

So I've got some Java code that uses Jakarta HttpClient like this:

URI aURI = new URI( "http://host/index.php?title=" + title + "&action=edit" );
GetMethod aRequest = new GetMethod( aURI.getEscapedPathQuery());

The problem is that if title includes any ampersands (&), they're considered parameter delimiters and the request goes screwy... and if I replace them with the URL-escaped equivalent %26, then this gets double-escaped by getEscapedPathQuery() into %2526.

I'm currently working around this by basically repairing the damage afterward:

URI aURI = new URI( "http://host/index.php?title=" + title.replace("&", "%26") + "&action=edit" );
GetMethod aRequest = new GetMethod( aURI.getEscapedPathQuery().replace("%2526", "%26"));

But there has to be a nicer way to do this, right? Note that the title can contain any number of unpredictable UTF-8 chars etc, so escaping everything else is a requirement.

+6  A: 

Here you go:

import java.net.URLEncoder;
...
...
URI aURI = new URI( "http://host/index.php?title=" + URLEncoder.encode(title,"UTF-8") + "&action=edit" );
GetMethod aRequest = new GetMethod( aURI.getPathQuery());

Check java.net.URLEncoder for more info.

Strelok
jpatokal
OK I fixed my answer. You shouldn't use .getEscapedPathQuery() because the title is now escaped by URLEncoder.
Strelok
Sorry, my brain was mush when I wrote my first reply -- you're right, it's enough to encode just parameter and leave the rest untouched. Thanks!
jpatokal
A: 

us the urlencoder class

yamspog
+1  A: 

Why are you calling getEscapedPathQuery() if you don't want the escaping? Just decide who's responsibility it is and be consistent.

Julian Reschke