PHP + jQuery：ajax回调不起作用

Question

I need your help to try to sort out an issue with ajax callback. What happen is that the php script is called without issues, the query works fine and it generate the json output, but the callback doesn't works. No div is displayed on success and no changes on class happens.

This is my form

<form class="form-inline">
<input type="hidden" id="id_pro" value="1">
<input type="hidden" id="status" value="1">
<button id="submit" type="button" class="btn btn-link">
<span id="check" class="glyphicon glyphicon-check" title="my tytle"></span>
</button>
</form>
<span class="error" style="display:none"> Please Enter Valid Data</span>
<span class="success" style="display:none"> Form Submitted Success</span>

This is the js part

$(function() {
 $("#submit").click(function() {
   var id_pro = $("#id_pro").val();
   var status = $("#status").val();
   var dataString = 'id_pro='+ id_pro + '&status=' + status;

   if(id_pro=='' || status=='')
   {
      $('.success').fadeOut(200).hide();
      $('.error').fadeOut(200).show();
   }
   else
   {
      $.ajax({
      type: "POST",
      url: "myphppage.php",
      data: dataString,
      datatype: 'json',
      success: function(data)
      {
          if(data.result=='1')
          {
            $('.success').fadeIn(200).show();
            $('.error').fadeOut(200).hide();
            $("#check").attr('class', data.check);
          }
       }
      });
    }
return false;
});
});

And this is the php part

<? 

    if($_POST)
    {
        $id_pro=$_POST['id_pro'];
        $status=$_POST['status'];
        if($status==0){$status=1;}else{$status=0;}
        if ($mysqli->query("UPDATE mytable SET online=".$status." WHERE id=".$id_pro." ") === TRUE) {
            header("Content-type: application/json");
            $data = array('check'=>'new_class','check_text'=>'new text','result'=>'1');
            print json_encode($data);   
        }
        else
        {
            header("Content-type: application/json");
            $data = array('result'=>'0');
            print json_encode($data);   
        }
        $result->close();
    }else { }
?>

Any idea? Thank you in advance

Answer 1

error 500 means error in php and in your php don't see defined $mysqli and $result i think here is your problem.

better PHP looks like this but must define connect to DB

<?php
header("Content-type: application/json");
$data = array('result'=>'0');

if ($_SERVER['REQUEST_METHOD'] == 'post' )
{
    $id_pro = $_POST['id_pro'];
    $status = ($_POST['status'] == 0) ? 1 : 0; // if($status==0){$status=1;}else{$status=0;}

    // define $mysqli
    if ($mysqli->query("UPDATE mytable SET online=".$status." WHERE id=".$id_pro." ") === TRUE) {
        $data = array('check'=>'new_class','check_text'=>'new text','result'=>'1');
    }
    // $result->close(); // ????
}

print json_encode($data);