Java Android Illegal Argument exception

Question

I'm executing the following code:

HttpClient httpclient = new DefaultHttpClient();

HttpResponse response = httpclient.execute(new HttpGet("(...)&avatarUrl={(...)}&socialId=1&sexo=m&username="));

My problem is I'm getting an illegalArgumentException at letter l from the word avatarUrl, and I don't understand why.

I'll really appreciate your help.

The fisrt parameter starts with an ?. I used (...) to shorten the url — user1867620, Jan 10 '13 at 20:33
If the suggestions answers below do not work out, please do post the full URL (but maybe with a different domain name). — praseodym, Jan 10 '13 at 20:35
http://bn.com/DarPerfasdfil?email=&id=5078789059&socialId=1&avatarUrl={"data":{"is_silhouette":false,"url":"http:\/\/profile.ak.fbcdn.net\/hprofile-ak-prn1\/161422_507789789789059_666973_q.jpg"}}&sexo=m&username= — user1867620, Jan 10 '13 at 20:39
Are you sure you want to pass that whole string in JSON-syntax as your avatarUrl and no only the "url" part? Also, consider my complete answer below to avoid encoding issues. For instance, braces should be escaped as praseodym pointed out below. — saschoar, Jan 10 '13 at 21:11

saschoar · Answer 1 · 2013-01-10T20:51:04.110

Besides the question of ? oder & in your code, there are cleaner ways to pass parameters:

HttpClient httpclient = new DefaultHttpClient();
HttpGet request = new HttpGet(url);
BasicHttpParams params = new BasicHttpParams();

params.setParameter("avatarUrl","...")
      .setParameter("socialId","...")
      .setParameter("sexo","...")
      .setParameter("username","...");

request.setParams(params);

HttpResponse response = httpclient.execute(request);

(not tested, but should work)

Note: I heavily edited my original answer due to a little misunderstanding.

score 0 · Answer 2 · edited May 23 '17 at 11:48

0

Looking at the source code of HttpGet, this exception means that the URL is invalid:

The first query string parameter should be preceded with ?
{ and } are not valid characters for URLs and should be escaped

edited May 23 '17 at 11:48

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answered Jan 10 '13 at 20:33

praseodym

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Java Android Illegal Argument exception

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