Android Intent send request to a link

Question

How can I send a request from my application to a link with out opening it in a browser? I tried Intent But its open in a browser; "Android"

Intent intent = new Intent(Intent.ACTION_VIEW, uri);
startActivity(intent);

Answer 1

If you want to display it in the browser then use the following code

if (!url.startsWith("http://")
                    && !url.startsWith("https://"))
                url = "http://" + url;
            browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
startActivity(browserIntent);

But if you dont want to open a browser and display it in your app itself then use WebView. There is an excellent tutorial at this link.

Answer 2

You can use a normal Java methods to accomplish this.

Use an HttpURLConnection, set the headers and content and send the request.

URL url = new URL(urlString);
HttpURLConnection con = (HttpURLConnection) url.openConnection();
// Set properties for con

// Set headers on this connection if required

// add url form parameters
DataOutputStream ostream = null;
try {
    ostream = new DataOutputStream(con.getOutputStream());
    String requestContents;
    // Set the contents of the request

    if (requestContents != null) {
                    // write the contents to the stream
        ostream.writeBytes(requestContents);
    }

} catch (Exception e) {
    e.printStackTrace();
} finally {
    if (ostream != null) {
        ostream.flush();
        ostream.close();
    }
}

Answer 3

try it

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();  
String link ="url";  
DefaultHttpClient hc=new DefaultHttpClient();  
ResponseHandler <String> res=new BasicResponseHandler();  
HttpPost postMethod=new HttpPost(link);  
List nameValuePairs = new ArrayList();  
nameValuePairs.add(new BasicNameValuePair("action","XYZ" ));  
nameValuePairs.add(new BasicNameValuePair("udid","XYZ" ));    
nameValuePairs.add(new BasicNameValuePair("user", "XYZ"));  

try {
    postMethod.setEntity(new UrlEncodedFormEntity(nameValuePairs));
      String response=hc.execute(postMethod,res);  
      System.out.println("response is== "+response);
      Toast.makeText(this,response, Toast.LENGTH_SHORT);
} catch (Exception e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

Answer 4

class RequestTask extends AsyncTask<String, String, String>{

    @Override
    protected String doInBackground(String... uri) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        String responseString = null;
        try {
            response = httpclient.execute(new HttpGet(uri[0]));
            StatusLine statusLine = response.getStatusLine();
            if(statusLine.getStatusCode() == HttpStatus.SC_OK){
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                out.close();
                responseString = out.toString();
            } else{
                //Closes the connection.
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (ClientProtocolException e) {
            //TODO Handle problems..
        } catch (IOException e) {
            //TODO Handle problems..
        }
        return responseString;
    }

To execute this class:

new RequestTask().execute(insert your Http link );