检查用户是否以JQuery等级登录并警告您是否必须登录

Question

检查用户是否已登录，如果用户已登录并显示警告以感谢评分，如果用户未登录，请登录以评分

<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>

此查询使用此PHP波纹管以提醒用户他/她已成功评分。 我设法仅在用户使用PHP登录时才插入评分，而在他/她未登录时不插入评分。但是，如何根据用户是否登录来修复JQuery以用适当的消息提醒用户在或没有？

 // 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：

在两个地方回显json编码的数组

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

在jQuery的职位：删除d.trim(status)相反

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