如何使用PHP更新和重新计算MySQL中的$ _ post值和变量

我在使用PHP脚本中更新MySQL中的某些值时遇到了一些困难。当我尝试使用以下代码更新下表行(用户"鲍勃"已经输入值以计算其总分和百分比，但是他的Test3等级必须更新为100，而不是先前的值为90(：

 <html>
 <head></head>
 <body>
  <form class="form" action=""  method="post">
 <?php
   $mysqli = new mysqli('', '', '', '');
   if(isset($_POST['calculate'])) {
   $name = $_POST['name'];
   $test1 = $_POST['test1'];
   $test2 = $_POST['test2'];
   $test3 = $_POST['test3'];
   $obtained = ($test1 + $test2 + $test3);
   $total = 300;
   $percentage = round(($obtained/$total)*100);
   $result = mysqli_query($mysqli, "INSERT INTO table1 (name, test1, 
   test2, test3, totalobtained, totalmarks, percent)
   VALUES ('$name', '$test1', '$test2', '$test3', 
   '$obtained', '$total', '$percentage')");
   }
   $conn = mysqli_connect('', '', '', '');
   if (!$conn) {
   die("Connection failed: " . mysqli_connect_error());
   }
   $sql = "UPDATE table1 SET test3='100', totalobtained='$obtained', 
   percent='$percentage' WHERE name='Bob'";

   if (mysqli_query($conn, $sql)) {
     echo "Record updated successfully";
   } else {
     echo "Error updating record: " . mysqli_error($conn);
   }
   mysqli_close($conn);
   ?>
   <table>
    <tr>
     <th>Name of Student*:</th>
     <td><input type="text" name="name"></td>
    </tr>
    <tr>
     <th>Test 1*:</th>
     <td><input type="number" name="test1"></td>
    </tr>
    <tr>
     <th>Test 2*:</th>
     <td><input type="number" name="test2"></td>
    </tr>
    <tr>
     <th>Test 3*:</th>
     <td><input type="number" name="test3"></td>
    </tr>
    <tr>
     <th>Total Marks Obtained:</th>
     <td><?php if(isset($_POST['calculate'])) { echo "$obtained";}?>
     </td>
    </tr>
    <tr>
      <th>Total Marks:</th>
      <td><?php if(isset($_POST['calculate'])) { echo "$total";}?>
     </td>
    </tr>
    <tr>
     <th>Percentage:</th>
     <td><?php if(isset($_POST['calculate'])) { echo "$percentage", 
     '%';}?></td>
    </tr>
    <tr>
     <th><input type="submit" name="calculate" value="Calculate"/>
     </th>
    </tr>
   </table>
   </form>
   </body>
   </html>

它将测试分数的测试3从上一个90分数中更新为100，但是，它并没有提取先前的测试分数来重新计算所获得的总数和百分比。结果，它更新了总共获得的，百分比为0。有些帮助将不胜感激，因为我是MySQL和PHP的新手。谢谢！

上表：

+----+-------+-------+-------+-------+----------------+-------------+---------+
| id | name  | test1 | test2 | test3 | totalobtained  | totalmarks  | percent |
+----+-------+-------+-------+-------+----------------+-------------+---------+
|  7 | Bob   |  100  |   100 |  90   |        290     |      500    |      96 |
+----+-------+-------+-------+-------+----------------+-------------+---------+

更新的表和更新语句的表：

+----+-------+-------+-------+-------+----------------+-------------+---------+
| id | name  | test1 | test2 | test3 | totalobtained  | totalmarks  | percent |
+----+-------+-------+-------+-------+----------------+-------------+---------+
|  7 | Bob   |  100  |   100 |  100  |        0       |      500    |       0 |
+----+-------+-------+-------+-------+----------------+-------------+---------+

只是看这个我相信SQL应该是

$sql = "UPDATE table1 SET test3='100', totalobtained=$obtained, 
   percent=$percentage WHERE name='Bob'";

您不需要'

就像Alex Howansky指出这很容易受到SQL注入的影响。

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