用PHP计算平均评级系统

Question

我对如何计算平均评级系统感到有点困惑。用户完全可以提供费率并将存储在数据库中，如下图所示

因此，想法是averagestar的价值来自平均评级表。例如，prov @ gmail有4个输入速率，平均值= 4，来自16/4。 （总星数/总费率）

从下面显示的代码中，我认为我的逻辑是错误的。 UPDATE平均SQL无效。

$sql= "
INSERT INTO rating VALUES 
(DEFAULT,'$prov', '$star','$detail', '$user')
";

$sql2="
DELETE FROM orderform 
 WHERE confirmation = 4 
   AND provider_username = '$prov' 
   AND user_username = '$user' 
   AND confirmation = 4
";

$sql3= "
SELECT AVG(rating) 
  FROM rating 
 WHERE provider_username = '$prov'
";

if (mysqli_query($conn, $sql) && mysqli_query($conn, $sql2)) {
$result = mysqli_query($conn, $sql3);
while($row = mysqli_fetch_array($result)) {
$sql= "
UPDATE theprovider 
   SET averagerating = '$result' 
 WHERE provider_username = '$prov'";
if (mysqli_query($conn, $sql)) {
echo 'success';    
}
}

有帮助吗？

Answer 1

你不能SET averagerating='$result' - 你应该用$ row来更新。

尝试：

$sql= "INSERT INTO rating VALUES (DEFAULT,'$prov', '$star','$detail', '$user')";
$sql2="DELETE FROM orderform WHERE confirmation=4 AND provider_username='$prov' AND user_username='$user' AND confirmation=4";
$sql3= "SELECT AVG(rating) AS myAvg FROM rating WHERE provider_username='$prov' GROUP BY provider_username";

if (mysqli_query($conn, $sql) && mysqli_query($conn, $sql2)) {
$result = mysqli_query($conn, $sql3);
while($row = mysqli_fetch_array($result)) {
$sql= "UPDATE theprovider SET averagerating='".$row["myAvg"]."' WHERE provider_username='$prov'";
if (mysqli_query($conn, $sql)) {
echo 'success';    
}
}

编辑：添加了GROUP BY子句。

Answer 2

AVG是一个聚合函数，因此应与GROUP BY一起使用。

SELECT AVG(rating) FROM rating WHERE provider_username = 'prov@gmail.com' GROUP BY provider_username;

Answer 3

star average rating dynamic from database using php 


try : - this code 



Datbase  - 


<?php 

$sum=0;

$sel_sups = my_select("SELECT  SUM(Rating) AS sum,Count(UserId) AS User FROM tbl_rating WHERE  ProductId='".$ProductId."'");  //  hard coded in here value UserId And ProductId

    $result=json_decode($sel_sups);    
    
    if($result->response==1){ 
    
    foreach($result->result as $myresults){
       
     
    $value = $myresults->sum;

    $sum = $value;
   
    $avg = $sum/$myresults->User;
     
    $no_format = number_format(floor($avg*100)/100,2, '.', '');
    
    $percent = $avg*100/5;
    
     echo '<div class="ratings">
     
     <div class="empty-stars"></div>
     
     <div class="full-stars" style="width:'.$percent.'%">
     
     </div>
     
     </div>'.'<a href="#" data-toggle="modal" data-backdrop="static" data-target="#reviewmodel" style="background: #5F283D;padding:10px;color: #fff;font-size: 12px;margin-left:5%">Write a Review</a>';

     echo '<div> 
     
     <p style="font-size: 13px;color:#333;margin-left:1%;margin-top: 1%;">Average Rating '.$no_format.'</p>
     
     </div>';

    }     
    
    } 

   ?>

{{1}}

Answer 4

这是实现这一目标的简单干净的方法：

<?php if(!empty($answer->AggRating)){ // This is fetched from your db
   $rating = (int)$answer->AggRating; // Converting it to integer
   $img="";
   $i=1;
   while($i<=$rating){
     $img=$img."<i class='fa fa-star' style='color: yellow;'></i>"; // You can edit your own html star.
     $i=$i+1;
   for($i=$rating+1; $i<=5; $i++){ // This is for non-star rating 
     $img=$img."<i class='fa fa-star'></i>";
   }
   echo $img;
} ?>