Question

我有移动div并将位置插入数据库的代码，但它不起作用：

使用example.php

<html>
<head>
  <script src="//code.jquery.com/jquery-1.10.2.js"></script>
  <script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>

  <script type="text/javascript">
$(document).ready(function() {
$( "#move" ).draggable({ 
containment: 'parent',

  stop: function(event, ui) {
      var posx = ui.offset.left;
      var posy = ui.offset.top;


      $.ajax({
          type: "POST",
          url: "coor.php",
          data: {x: posx, y: posy}
        }).done(function( msg ) {
          alert( "Data Saved");
        }); 
  }
 });
});
</script>
</head>
<body>
 <div style="width: 400px; height: 500px; margin: 0 auto; border: 1px solid;">
<div id="move" style="width: 100px;">
<p>Move this text with your mouse holding click!!</p> 
</div></div>
</body>
</html>

另一个是coor.php

<?php   session_start();

    include_once('../connect.php');

    $probe = $_SESSION['logger'];




        $pos_x = $_POST['x'];

        $pos_y = $_POST['y'];

            $sql = mysqli_query($con, "SELECT * FROM probe WHERE fromid='$probe'");
            $result = mysqli_num_rows($sql);
            if($result == 1){


                $sqliupdate = "UPDATE probe SET x='$pos_x' y='$pos_y' WHERE fromid='$probe' LIMIT 1";
                if(mysqli_query($con,$sqliupdate)){

                    echo "Coords are updated!";

                }
            }else{

                $sqliinsert = "INSERT INTO probe (id, x, y, fromid) VALUES ('','$pos_x','$pos_y','$probe') LIMIT 1";
                if(mysqli_query($con,$sqliinsert)){

                    echo "Coords are inserted!";

                }               


            }



?>

我移动div后我得到了ajax msg ...但是它没有插入或更新mysql数据库....我做错了什么？

插入到db不能使用ajax和php

0 个答案: