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    <p>The string that's causing this error is <strong>the response</strong>, not the request one. Also, with your code <strong>you are not sending a <code>JSONObject</code></strong>, you are just sending 5 different parameters.</p> <p>Let's begin with the request. You cannot send a <code>JSONObject</code> directly to your server, you need to send it as a <code>String</code> and then parse it in the server to get the JSONObject back. The same goes with the response, you will receive a String to parse.</p> <p>So let's create a <code>JSONObject</code> in your client and add it to the parameters list:</p> <pre><code>// Let's build the JSONObject we want to send JSONObject inner_content = new JSONObject(); inner_content .put(CLASSICLevel, "1") .put(CURRENTLevel, "2") .put(UPDATEDate, "345") .put(NAME, "Nil") .put(UPDATETIME, "10"); JSONObject json_content= new JSONObject(); json_content.put("objTimesheet", inner_content); // TO PRINT THE DATA Log.d(TAG, json_content.toString()); // Now let's place it in the list of NameValuePair. // The parameter name is gonna be "json_data" List&lt;NameValuePair&gt; params1 = new ArrayList&lt;NameValuePair&gt;(); params1.add(new BasicNameValuePair("json_data", json_content.toString())); // Start the request function json = jparser.makeHttpRequest(url_login, "POST", params1); </code></pre> <p>Now your server will receive just <strong>ONE parameter called "objTimesheet"</strong> whose content will be a <code>String</code> with the json data. If your server script is <code>PHP</code>, you can get the JSON Object back like this:</p> <pre><code>$json = $_POST['json_data']; $json_replaced = str_replace('\"', '"', $json); $json_decoded = json_decode($json_replaced, true); </code></pre> <p>$json_decoded is an array containing your data. I.e., you can use $json_decoded["Name"].</p> <p>Let's go to the response now. If you want your client to receive a <code>JSONObject</code> you need to send a <strong>valid</strong> string containing a <code>JSONObject</code>, otherwise you get the <code>JSONException</code> you are getting now.</p> <p>The String: "InsertTimesheetItemResult=Inserted successfully" <strong>IS NOT a valid JSON string</strong>. </p> <p>It should be something like: "{"InsertTimesheetItemResult" : "Inserted successfully"}".</p> <p>PHP has the function <strong>json_encode</strong> to encode objects to JSON strings. To return a String like the one I wrote above, you should do something like this:</p> <pre><code>$return_data["InsertTimesheetItemResult"] = "Inserted successfully"; echo json_encode($return_data); </code></pre> <p>I hope this helps you.</p>
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    1. POGetting Error in JSON Parsing while sending an JSON object with parameter
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    1. USPX Developer
    1. USPX Developer
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    1. COBut the objTimesheet is an JSON object i want to send this also with the makeHttpRequest.
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    2. COIn my code above I'm sending a parameter named "objTimesheet" whose value is: "{"ClassicLevel":"1", "CurrentLevel":"2", "UpdatedDate":"345", "Name":"Nil", "UpdatedTime": "Nil"}". That's not what you want?
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    3. COString is:-{"UpdatedTime":"10","Name":"Nil","CurrentLevel":"2","Message":"Hi","UpdatedDate":"345","ClassicLevel":"1"}
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