CODEWITHSUNDEEP
javaandroidurluri
Marcos Bergamo
Marcos Bergamo

Reputation: 1123

URL Illegal Character

Here is my code:

HttpClient client = new DefaultHttpClient();
            client.getParams().setParameter(CoreProtocolPNames.USER_AGENT, "android");
            HttpGet request = new HttpGet();
            request.setHeader("Content-Type", "text/plain; charset=utf-8");
            Log.d("URL", convertURL(URL));
            request.setURI(new URI(URL));
            HttpResponse response = client.execute(request);
            bufferedReader = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
            StringBuffer stringBuffer = new StringBuffer("");
            String line = "";
            String NL = System.getProperty("line.separator");

I don't know which error in my URL:

http://localhost/CyborgService/chatservice.php?action=recive_game&nick_sender=mkdarkness&pass=MV030595&date_last=2012-11-18 09:46:37&id_game=1

I have already used a function to convert URL, but has not worked. But, if I trying open this URL in my Browser, it opens successfully.

Here is my error:

11-18 21:46:37.766: E/GetHttp(823): java.net.URISyntaxException: Illegal character in query at index 127: http://192.168.0.182/CyborgService/chatservice.php?action=recive_game&nick_sender=mkdarkness&pass=MV030595&date_last=2012-11-18 09:46:37&id_game=1

Upvotes: 3

Views: 29946

Answers (3)

user2719792
user2719792

Reputation:

Do this way it will definetly help you

    HttpClient myClient = new DefaultHttpClient();

            HttpPost myConnection = new HttpPost("http://192.168.1.2/AndroidApp/SendMessage");

            try {

    //Your parameter should be as..

                List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
                nameValuePairs.add(new BasicNameValuePair("messageText", msgText));
                nameValuePairs.add(new BasicNameValuePair("senderUserInfoId", loginUserInfoId));

//set parameters to ur URL

                myConnection.setEntity(new UrlEncodedFormEntity(nameValuePairs));
//execute the connection
                HttpResponse response = myClient.execute(myConnection);
    }
    catch (ClientProtocolException e) {

                //e.printStackTrace();
            } catch (IOException e) {
                //e.printStackTrace();
            }

Upvotes: 0

alexfernandez
alexfernandez

Reputation: 1988

There is a space in your URL, in position 127. The date is generated as "date_last=2012-11-18 09:46:37", which causes an error when opening the URL.

Spaces are not formally accepted in URLs, although your browser will happily convert it to "%20" or to "+", both valid representations of a space in a URL. You should escape all characters: you can replace space with "+" or just pass the String through URLEncoder and be done with it.

To use URLEncoder see e.g. this question: encode with URLEncoder only parameter values, not the full URL. Or use one of the constructors for URI which have a few parameters, not a single one. You are not showing the code that constructs the URL so I cannot comment on it explicitly. But if you have a map of parameters parameterMap it would be something like:

String url = baseUrl + "?";
for (String key : parameterMap.keys())
{
  String value = parameterMap.get(key);
  String encoded = URLEncoder.encode(value, "UTF-8");
  url += key + "&" + encoded;
}

Some other day we can talk about why Java requires to set the encoding and then requires that the encoding be "UTF-8", instead of just using "UTF-8" as the default encoding, but for now this code should do the trick.

Upvotes: 10

jlordo
jlordo

Reputation: 37853

There is a whitespace character:

...2012-11-18 09:46:37... (at index 127, just like the error message says).

Try replacing it with %20

Upvotes: 2

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