Ajax Jquery ติดการส่งข้อมูลไปแบบไม่ให้รีเฟรชหน้าจอ

<?php
	include('connection/connection.php');
?>
<?php 
	if($_GET["work"]=='N') {
			$strSQL = "UPDATE work_descrip SET working = 'Y' WHERE work_id = '".$_GET["id"]."' ";
			echo 'OK';
		}
	else {
			echo 'No';
	}
	$objQuery = mysql_query($strSQL) or die(mysql_error());
?>

<?php 
                      if ($objResult["working"] == "Y") {
                        $work = "img/ico_bifag.png";
                      }
                      else {
                        $work = "img/no.png";
                      }
                    ?>
                    <td align="center">
                       
                       <a href="#" id="content" class="submit">
                          
                            <img src="<?=$work ?>" class="work">
                         
                        </a>
                        <span id="status"></span>
                    </td>

<script type="text/javascript">
  $(document).ready(function()
{
  $(".submit").click(function() 
{
  //var txt = $('#form1').serialize(); 
  var msgbox = $("#status");
  var content = $("#content").click();

  if(content)
{
$("#status").html('<img src="img/ajax-load.gif" align="absmiddle">&nbsp;Checking availability...');

$.ajax({  
    type: "POST",  
    url: "update_work.php",  
    data: { 'work' : 'N', 'work_id' : 1},  //ค่าจาก input ใน form1     1ต้องมาจากการคิวรี่
    success: function(msg){  
   $("#status").ajaxComplete(function(event, request){ 
  if(msg == 'OK')
  { 
  
      $("#status").show();
      msgbox.html('<img src="img/ico_bifag.png" align="absmiddle">');
  }  
   else  
  { 
     $("#status").show();
     msgbox.html('<img src="img/no.png" align="absmiddle">');
  }  
});
}

});
}
});
  });
</script>