java.lang.IllegalArgumentException：Android中索引77的查询中的非法字符

Question

I am trying to parse Facebook feed using json

it show this error

06-02 16:53:33.112: D/ee:(29180): java.lang.IllegalArgumentException: Illegal character in query at index 77: https://graph.facebook.com/331394590231184/feed?access_token= ******&client_id=**&client_secret=*****?

i use code :

private static String url = "https://graph.facebook.com/331394590231184/feed?access_token=**|*****&client_id=***&client_secret=****";
JSONParser jParser = new JSONParser();
List<NameValuePair> params = new ArrayList<NameValuePair>();
JSONObject json = jParser.makeHttpRequest(url, "GET", params);
JSONArray data = json.getJSONArray("data");

JSONParser code :

static InputStream is = null;
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();

Answer 1

You're building an invalid URL with multiple ? in it, you should pass just the scheme, host, and path as the url variable, and then pass the params separately:

private static String url = "https://graph.facebook.com/331394590231184/feed";
JSONParser jParser = new JSONParser();
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("access_token", "**|*****"));
params.add(new BasicNameValuePair("client_id", "***"));
params.add(new BasicNameValuePair("client_secret", "****"));
JSONObject json = jParser.makeHttpRequest(url, "GET", params);
JSONArray data = json.getJSONArray("data");

Another way to fix it, which you should probably do anyway, is to make sure that params has anything in it prior to processing it in your JSONParser code:

static InputStream is = null;
DefaultHttpClient httpClient = new DefaultHttpClient();
if (params != null && params.size() > 0){
    String paramString = URLEncodedUtils.format(params, "utf-8");
    url += "?" + paramString;
}
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();

Answer 2

https://www.google.com/search?q=valid+url+characters -- do you see anything unusual around character 77? Also, https://www.google.com/search?q=encode+url

BTW, editing a question doesn't remove a previous revision: https://stackoverflow.com/posts/30601529/revisions