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In xampp my code is working fine but when i uploaded on hosting account it gives empty output json array.i need to get data using t.user_id_sender = '$userdata'. when i give index.php?user=2 i want to show data in json format where user_id_sender = 2 and all user_id_receiver shows only one time not repeated like sender id:2
receiver id:2,3,4,5,6 but sender id 2 should not included in receiver output should be
receiver id:3,4,5,6

  <?php

$userdata=$_GET["user"];

$username="root";
$password="";
$host="localhost";
$db_name="chat";
$con=mysqli_connect($host,$username,$password,$db_name);  
$sql="SELECT t.*, p.name FROM nearby t INNER JOIN users p ON          
t.user_id_receiver = p.user_id WHERE t.user_id_sender = '$userdata' group by 
t.user_id_receiver";                              
$result=mysqli_query($con,$sql);
$response=array();

   while($row= mysqli_fetch_array($result))

    {
        $cmt = array();
        $cmt["message"] = $row["message"];
        $cmt["created_at"] = $row["created_at"];
        $cmt["message_id"] = $row["message_id"];
        $cmt["name"] = $row["name"];
        $cmt["user_id_sender"] = $row["user_id_sender"];
        $cmt["user_id_receiver"] = $row["user_id_receiver"];


       array_push($response,$cmt);
      }
     echo json_encode($response);
      ?>

1 Answers1

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This might be problem with your database connection ,try to debug it, print_r($cmt) at the end of while loop and if no data recevied then correct your database connection. your code is good and have no issue

Aammad Ullah
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