Question

这是我用来尝试更新记录的代码。我没有看到任何问题，但它没有引发错误并指出运行成功。我尝试了几种其他方法，它们只是不更新MSSQL表中的行。数据库/表在运行Windows 10和IIS的我的机器上是本地的（我知道！）。

<?php

/**
 * Use an HTML form to edit an entry in the
 * users table.
 *
 */

 require "C:\inetpub\wwwroot\Remediation\config.php";
 require "C:\inetpub\wwwroot\Remediation\common.php";

 if (isset($_POST['submit'])) {
   if (!hash_equals($_SESSION['csrf'], $_POST['csrf'])) die();

   try {
     $connection = new PDO($dsn, $username, $password);

     $user =[
       "id"        => $_POST['id'],
       "firstname" => $_POST['firstname'],
       "lastname"  => $_POST['lastname'],
       "email"     => $_POST['email'],
       "age"       => $_POST['age'],
       "location"  => $_POST['location'],
       "date"      => $_POST['date']
     ];

     $sql = "UPDATE users
             SET id = :id,
               firstname = :firstname,
               lastname = :lastname,
               email = :email,
               age = :age,
               location = :location,
               date = :date
             WHERE id = :id";

   $statement = $connection->prepare($sql);
   $statement->execute($user);
   } catch(PDOException $error) {
       echo $sql . "<br>" . $error->getMessage();
   }
 }

 if (isset($_GET['id'])) {
   try {
     $connection = new PDO($dsn, $username, $password);
     $id = $_GET['id'];
     $sql = "SELECT * FROM users WHERE id = :id";
     $statement = $connection->prepare($sql);
     $statement->bindValue(':id', $id);
     $statement->execute();

     $user = $statement->fetch(PDO::FETCH_ASSOC);
   } catch(PDOException $error) {
       echo $sql . "<br>" . $error->getMessage();
   }
 } else {
     echo "Something went wrong!";
     exit;
 }
 ?>    

 <?php if (isset($_POST['submit']) && $statement) : ?>
    <blockquote><?php echo escape($_POST['firstname']); ?> successfully 
 updated.</blockquote>
 <?php endif; ?>

 <h2>Edit a user</h2>

 <form method="post">
     <input name="csrf" type="hidden" value="<?php echo 
   escape($_SESSION['csrf']); ?>">
     <?php foreach ($user as $key => $value) : ?>
       <label for="<?php echo $key; ?>"><?php echo ucfirst($key); ?></label>
        <input type="text" name="<?php echo $key; ?>" id="<?php echo $key; ?>" value="<?php echo escape($value); ?>" <?php echo ($key === 'id' ? 'readonly' : null); ?>>
     <?php endforeach; ?>
     <input type="submit" name="submit" value="Submit">
 </form>

我到底在哪里错。我可以连接，修改和创建就好了。我是否需要调整$ USER或查询？

Answer 1

您是否直接在MSSQL Management Studio中尝试了该查询？

之后，您会看到它工作正常，请尝试“回显”您的POST变量，以查看您是否正确获取了它们。

Answer 2

我犯了一个菜鸟错误，我试图在MSSQL中更新ID和时间戳列，这是不允许的。我将自己的代码更改为此，并且效果很好：

   $user =[
     "id"        => $_POST['id'],
     "firstname" => $_POST['firstname'],
     "lastname"  => $_POST['lastname'],
     "email"     => $_POST['email'],
     "age"       => $_POST['age'],
     "location"  => $_POST['location'],
           ];

   $sql = "UPDATE users
           SET firstname = :firstname,
             lastname = :lastname,
             email = :email,
             age = :age,
            location = :location
           WHERE id = :id";

Answer 3

我没有看到任何INSERT。相反，您仅使用UPDATE，这意味着仅修改现有记录。这是您想要的吗？

 $sql = "UPDATE users
         SET id = :id,
           firstname = :firstname,
           lastname = :lastname,
           email = :email,
           age = :age,
           location = :location,
           date = :date
         WHERE id = :id";

您在这里选择id = :id的（全部）记录，然后在SET id = :id处将id设置为一个已设置的值。 Is不会破坏某些内容，但这暗示您逻辑中的某些内容是错误的。

顺便说一句，您允许每个客户端更改记录。不是SQL注入，但很可能非常糟糕。

无法通过php更新sql表

3 个答案: