Question

我是初学程序员，我正在尝试查找供应商2的平均用户评级我的步骤如下：
1.用户'Jimmy'给供应商2 评分3 （评分超过5）
2.TABLE vendoratings 已更新
3. ratingstot.php 然后在<所有供应商之前计算评级和响应数量的总和< strong>更新下面显示的表 vendortotalratings

4.User'Jimmy'点击'查看平均用户评分'
5.检索 totalratings 和 totalno 的值并将其划分为Javascript

var average= totalratings/totalno

6。 totalno将显示给用户Jimmy。结束

问题
1.我需要帮助在ratingstot.php中形成for或while循环来计算评级和否。属于供应商X的响应，然后在每个供应商的 vendortotalratings 表中插入它们 ratingstot.php

<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");
error_reporting(E_ERROR);
try{
    //Database connection
    $conn = new mysqli("localhost", "XXXXXXXX_XXX", "XXXXXXXX", "XXXXXXXX");

    //Unsure how to loop this to make vendor new value every loop
    for($i=0; $i<=6; $i++){
    $vendor = ??

        //Calculate sum of ratings from table ratings
        $result = $conn->query("SELECT SUM(ratings) FROM ratings WHERE vendorid = '".$vendorid."' ");
        $row = mysqli_fetch_array($result);
        $totalratings = $row[0];

        //Calculate no. of responses (by counting no. of rows)
        $result1 = $conn->query("SELECT * FROM ratings WHERE vendorid = '" . $vendorid."' ");
        $totalno = mysqli_num_rows($result);

        //inserting the results into the table
        $query  = " UPDATE vendortotalratings SET ";
        $query .= " totalratings = '". $totalratings ."', totalno='".$totalno."' ";
        $query .= " WHERE vendorid = '". $vendorid ."'";
        $result2 = $conn->query($query);

    }
    echo($outp);
}

catch(Exception $e) {
    $json_out =  "[".json_encode(array("result"=>0))."]";
    echo $json_out; 
}
?>

我不知道如何循环这个，是否有更简单的步骤来计算每个供应商的平均评级？

Answer 1

该问题并未阐明如何首先在vendortotalratings表中插入记录。因此，假设此表中已存在每个供应商的记录，则不必编写全新的循环。

更新vendortotalratings：
SQL可以在单个查询中计算总评级及其计数，然后可以替换您拥有的循环。

UPDATE vendortotalratings vtr
       INNER JOIN 
       (    
         SELECT vendorid, SUM(ratings) AS sumratings, COUNT(ratings) AS countratings 
                FROM vendoratings 
         GROUP BY vendorid
       ) vr
    ON  vtr.vendorid    =   vr.vendorid
    SET 
       vtr.totalratings =   vtr.totalratings + vr.sumratings
      ,vtr.totalno      =   vtr.totalno      + vr.countratings

计算平均值：
至于你的第二个问题，要计算平均值，你可以运行以下查询，它将为你提供运行时结果：

SELECT vendorid, totalratings, totalno, 
       CAST((totalratings/totalno) AS DECIMAL(10, 2)) AS avgrating 
FROM vendortotalratings;

如果您从结果中获取关联数组，或者使用适当的索引号（在本例中应为{），则可以使用avgrating直接在PHP中访问变量$row['avgrating'] {1}}

Answer 2

您可以像这样简化所需的解决方案，而不是所有这些复杂的事情：（假设：我假设 vendorid ，评级， totalratings 和 totalno 列的类型为{{1 }}）

使用以下声明/查询获取与每个 vendorid 对应的总计和 totalno 。
```
INT
```
现在使用$result = $conn->query("SELECT vendorid, SUM(ratings) as totalratings, COUNT(userid) as totalno FROM vendorratings GROUP BY vendorid");循环遍历$result结果集。
```
while()
```
在上述while($row = $result->fetch_array()){ ... }循环的每次迭代中，检查 vendorid 值是否已存在。如果存在，则while()行包含总计和 totalno ，否则UPDATE新行包含 vendorid ，总计和 totalno 。
```
INSERT
```

{strong} while($row = $result->fetch_array()){ $res = $conn->query("SELECT vendorid FROM vendortotalratings WHERE vendorid = " . $row['vendorid']); if($res->num_rows){ // Update the existing row $conn->query("UPDATE vendortotalratings SET totalratings = ".$row['totalratings'].", totalno = ".$row['totalno']." WHERE vendorid = ".$row['vendorid']); echo "Affected rows: " . $conn->affected_rows . '<br />'; }else{ // Insert a new row $res = $conn->query("INSERT INTO vendortotalratings VALUES(".$row['vendorid'].", ".$row['totalratings'].",".$row['totalno'].")"); if($res) echo "New row inserted <br />"; } }块的完整代码就像这样：

try-catch

如何循环表中的每个值并找到平均值

2 个答案: