Question

Why this json encode format is not working .If I write it outside of the all brackets.Then it works,but get only one row.Plz help

$sql=mysqli_query($database,"SELECT * FROM com  ORDER by time DESC");
while($row=mysqli_fetch_array($sql)){
        $mid=$row['id'];
        $text=$row['text'];

        if($text) {
            $response=array();

            $response['msg']=$text;
            $response['id']=$id;
            echo json_encode($response);
        }
}
    //$response=array();

    // $response['msg']=$text;
    // $response['id']=$mid;
    // echo json_encode($response);

Jquery for getting above results.

function com(id){
    $.post('load.php', {tocom:id},function(data) {
        var json = eval('(' + data + ')');
        if(json['msg'] != "") {
            var msg=json['msg'];
            var fro=json['id'];
            $("#msg).html(fro+msg);
        }
      });
}

Answer 1

You are not giving one json response, but as much response as you have iterations in your while.

The good syntax should be :

PHP :

$sql=mysqli_query($database,"SELECT * FROM com  ORDER by time DESC");
$final_response = array();
while($row=mysqli_fetch_array($sql)){
    $mid=$row['id'];
    $text=$row['text'];

    if($text) {
        $response=array();
        $response['msg']=$text;
        $response['id']=$id;
        $final_response[] = $response;
    }

}
echo json_encode($final_response);

JS :

function com(id){

    $.post('load.php', {tocom:id},function(data) {
        $("#msg").html('');
        var json = eval('(' + data + ')');
        $.each(json, function(i, row) {
           if(row['msg'] != "") {
               var msg=row['msg'];
               var fro=row['fro'];
               $("#msg").append(fro+msg);
           }
        });
    });

}