php - 使用 php 和 mysql 更新某些用户

Question

这个问题已经有答案了:

"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP (29 个回答)

已关闭 3 年前。

我在进入一些用户的个人资料后尝试更新他们的信息。

这是将我转到他们的个人资料页面的按钮。一切正常，直到我尝试更新一些信息。

echo "<td>" .'<a  href="profile.php?id=' . $row['id']. '"><button class="view">View Profile</button></a>'. "</td>";

这是个人资料页面:

  $id = $_GET['id'];
  $db = mysqli_connect('localhost', '###', '###', '###');
  $sql = "SELECT * FROM clients  WHERE id='".$id."'  ";

  $result = mysqli_query($db, $sql);

    $resultCheck = mysqli_num_rows($result);

    if ($resultCheck > 0) {

    echo "<table>
            <tr>
                <th> Naam </th>
                <th> id </th>
                <th> email </th>

            </tr>";
        while ($row = mysqli_fetch_assoc($result)) {
      echo "<tr>";
            echo  "<td>" . $row['name'] . "</td>";
            echo  "<td>" . $row['id'] . "</td>";
            echo "<td>" . $row['email']  . "</td>";   

            $name = $row['name'];
            $email = $row['email'];

    }
  }  

    if (isset($_POST['pinfo'])) {

      $query = "UPDATE clients SET name=? , email=?
       WHERE id=?";
      $stmt = mysqli_prepare($db, $query);
      mysqli_stmt_bind_param($stmt, 'sss',   $_POST['name'], $_POST['email'], $id);
      mysqli_stmt_execute($stmt);
   }

尝试更新后，我收到错误 "Undefined index: id"

Answer 1

您的 id 字段未初始化或由于某种原因处于休息状态。尝试通过将 id 设置为隐藏的表单元素来传递表单，例如:

<input stype="diplay: none;" name="id" value='"+$id+"' />

并将其从 $_POST 传递给 mysql_stmt_bind_param

 mysqli_stmt_bind_param($stmt, 'sss',   $_POST['name'], $_POST['email'], $_POST['id']);