Web API httpget with many parameters

Web API httpget with many parameters

By : CH4TTchitto
Date : November 29 2020, 01:01 AM
hope this fix your issue I ended up passing extra parameters with my httpget call. I will probably follow this pattern unless I get some additional feedback.
code :

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Httpget removes parameters

Httpget removes parameters

By : MvN
Date : March 29 2020, 07:55 AM
around this issue The mistake was from the server, when requesting this URL, it removes the header JSON, so you need to add : httpget.addHeader(new BasicHeader("Accept", "application/json")); under the HttpGet
here is the full code:
code :
 HttpClient httpclient = new DefaultHttpClient();

    // Prepare a request object
    HttpGet httpget = new HttpGet(url);
    httpget.addHeader(new BasicHeader("Accept", "application/json"));
    //httpget.getParams().setParameter("format", "JSON");


    // Execute the request
    HttpResponse response;

    String result = null;
    try {
        response = httpclient.execute(httpget);

        // Get hold of the response entity
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        // to worry about connection release

        if (entity != null) {
            // A Simple Response Read
            InputStream instream = entity.getContent();
            result = convertStreamToString(instream);
            // Closing the input stream will trigger connection release
    } catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
    } catch (IOException e) {
        // TODO Auto-generated catch block

    return result;

private static String convertStreamToString(InputStream is) {
    * To convert the InputStream to String we use the BufferedReader.readLine()
    * method. We iterate until the BufferedReader return null which means
    * there's no more data to read. Each line will appended to a StringBuilder
    * and returned as String.
    BufferedReader reader = new BufferedReader(new InputStreamReader(is), 8192);
    StringBuilder sb = new StringBuilder();

    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
    } catch (IOException e) {
    } finally {
        try {
        } catch (IOException e) {

    return sb.toString();
Android - HttpGet with parameters

Android - HttpGet with parameters

By : sagar reddy
Date : March 29 2020, 07:55 AM
will help you You do not need the JSONArray (since there is no square brackets in the responce)! Please try instead of
code :
      JSONArray respJSON = new JSONArray(respStr);
      for (int i = 0; i < respJSON.length(); i++) {
           JSONObject obj = respJSON.getJSONObject(i);
           String name = obj.getString("name");
      JSONObject respJSON = new JSONObject(respStr);
      String name = respJSON.getString("name");
HttpGet with parameters using HttpWebResponse C#

HttpGet with parameters using HttpWebResponse C#

By : Lucas Black Pearl
Date : March 29 2020, 07:55 AM
should help you out Try using URL encoding.
Multiple Parameters to httpget function

Multiple Parameters to httpget function

By : user2882023
Date : March 29 2020, 07:55 AM
hop of those help? The only thing you need is to add another parameter in your action method:
code :
public ActionResult Details(string id, string anotherParameter)
java.lang.NullPointerException at HttpGet httpGet = new HttpGet(url);

java.lang.NullPointerException at HttpGet httpGet = new HttpGet(url);

By : Lena
Date : March 29 2020, 07:55 AM
Does that help My guess would be that you are either passing a NULL string into your method, or for some reason the string is getting set to NULL if the HttpGet() method fails to resolve the URL or something else goes wrong. Try appending this at the end of your try / catch statement:
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