檢查用戶是否以JQuery等級登錄並警告您是否必須登錄

Question

檢查用戶是否已登錄，如果用戶已登錄並顯示警告以感謝評分，如果用戶未登錄，請登錄以評分

<script>
$(document).ready(function () {
   $("#demo2 .stars").click(function ()  {

        $.post( "rate.php",{rate:$(this).val(), vid_id : 
       $vid_result['vid_id']}, function(d) {
        console.log(d);
        //JSON.parse(d);

         if(d.status && d.trim(status) == 'active'){

          alert('Thanks For Rating');

          }
          else if(d.status && d.trim(status) == 'inactive'){

          alert('You have to be logged in to rate'); 

         }

        },"json");
      });
    });
 </script>

此查詢使用此PHP波紋管以提醒用戶他/她已成功評分。 我設法僅在用戶使用PHP登錄時才插入評分，而在他/她未登錄時不插入評分。但是，如何根據用戶是否登錄來修復JQuery以用適當的消息提醒用戶在或沒有？

 // this page is rate.php

 <?php

  include 'core/init.php'; 

 if (isset($_POST['rate']) && !empty($_POST['rate'])) {
  if(isset($session_user_id)){
  $rate = sanitize($_POST['rate']);
   $vidid = sanitize($_POST['vid_id']);
 // check if user has already rated
   $sql = "SELECT `id` FROM `rating` WHERE `rater_id`='" . 
  $session_user_id."' AND  `video_id` = '".$vidid."'";
  $result = $conn->query($sql);
  $row = $result->fetch_assoc();
  if ($result->num_rows > 0) {

     $sql = "UPDATE rating SET `rate` = '$rate' WHERE `rater_id`='" . $session_user_id . "' AND `video_id` = '".$vidid."'";
      if (mysqli_query($conn, $sql)) {        }
    echo $row['id'];
} else {

    $sql = "INSERT INTO `rating` (`video_id`, `rater_id`, `rate`) VALUES ('$vidid' ,'$session_user_id', '$rate')";
    if (mysqli_query($conn, $sql)) {
     //success
    json_encode(array('status'=>'active'));
    }else
    echo "fail";
}
}else{
     //not logged in
     json_encode(array('status'=>'inactive'));
  }
  }

?>

Answer 1

rate.php：

在兩個地方回顯json編碼的數組

echo json_encode(array('status'=>'active'));

在jQuery的職位：刪除d.trim(status)相反

d.status && d.status == 'active'