总和MySQL特定的表列并将结果写入其他列

Question

我通过财政的名字有一个MySQL表，并有ID，平衡，要求，ytotal6和预算列。 我要总结的余额列值和ytotal6结果写入其中id是1列。

这是表结构link

这是我的index.php文件

$con=mysqli_connect("127.0.0.1","root","","dji001"); 
       $result = mysqli_query($con,"SELECT * FROM finance") 
       or die("Error: ".mysqli_error($con)); 

       echo "<form style='width:1080px; backgrond-color:transparent;' action='update.php' method='post' class='form-group'>"; 
        while($row = mysqli_fetch_array($result)) 
        { 
        $id = $row['ID']; 
        $Budget = $row['Budget']; 
        $Balance = $row['Balance']; 
        $Requested = $row['Requested']; 
        $balance = $Requested + $Budget; 

        $con->query("UPDATE finance SET balance = $balance WHERE id = $id"); 
        echo "<div class='calc_container'> 

        <input type='hidden' class='id' name='id[]' value='".$row['ID']."'> 

        <input type='text' class='budget' name='Budget[]' value='".$row['Budget']."'> 

        <input type='text' class='req_kbl' name='Requested[]' value='".$row['Requested']."'> 

        <input type='text' class='balance' name='Balance[]' value='".$row['Balance']."'> 

        </div>";} 

          $result = mysqli_query($con,"SELECT * FROM finance where id=1") 
          or die("Error: ".mysqli_error($con)); 
          while($row = mysqli_fetch_array($result)) 
          echo" 
          <input type='text' class='ytotal6' name='ytotal6' value='".$row['ytotal6'] ."'>"; 
          echo "<input type='Submit' class='button' value='Submit'></form>"; 

我试图$ytotal6 = sum[$balance];，但没有奏效。

Answer 1

对于“总结余额列值和结果写在ytotal6其中id是1列”试试这个

UPDATE finance f1, (SELECT SUM(balance) AS bal 
      FROM finance) f2 
    SET ytotal6 = bal 
    WHERE f1.id = 1; 

注：对于错误，如“你不能指定目标表中更新FROM子句”我们需要创建一个临时表