jquery - 每秒检查数据库值

Question

我遇到了jquery的问题。加载php页面时不会触发下面的代码。我试图检查数据库表中字段的值，并在返回值为0时进行重定向。我已经花了好几个小时，我已经真正搜索到了找到答案。任何帮助表示赞赏。

<script type="text/javascript">
$jQn(document).ready(function(){
var do_checking;
var status = 1;
var ui = "";

        if(status == 1){
        $jQn.blockUI.defaults.overlayCSS.opacity=.3;
                    $jQn.blockUI({
                        message: ui,
                        css: {
                            border: '10px solid #888',
                            backgroundColor: '#000',
                            '-webkit-border-radius': '10px',
                            '-moz-border-radius': '10px',
                            color: '#fff',
                            width: '50%',
                            left: '25%',
                            cursor: 'wait',
                            '-ms-filter': 'progid:DXImageTransform.Microsoft.Alpha(Opacity=10)',
                            'filter': 'progid:DXImageTransform.Microsoft.Alpha(Opacity=10)',
                            '-moz-opacity':.10,
                            opacity:.10,
                            padding:'15px'
                        }
                    });
            do_checking = setInterval(check_search_status, 1000);
        } else {
            document.location = 'https://mysite.com/result.php';
        }

});

function check_search_status(){
    $jQn.ajax({
        type: 'POST',
        url: 'https://mysite.com/checksearchstatus.php',
        data: {
            id: '$owner'
        },
        dataType: 'json',
        success: function(response){
            if(response.status == 0){
                clearInterval(do_checking);
                document.location = 'https://mysite.com/result.php';
            }
        }
    });
}
</script>

这是分离的php文件的代码

$con = mysql_connect($db_host,$db_username,$db_password);

if (!$con)
  {
  die('Feil ved tilkobling av: ' . mysql_error());
  }

mysql_select_db($db_name, $con);


$response = array('status' => '');

$user_id = $_POST['id'];

// Build SQL
$first_search_sql = "SELECT `firstsearch` FROM `class_users` WHERE id = '".$user_id."'";

$first_search_result = $db->query($first_search_sql);

// Fetch result
$first_search_row = $first_search_result->fetch();

// Free result
$first_search_result->freeResult();

$response['status'] = $first_search_row['firstsearch'] ;

echo json_encode($response);