這是我用來嘗試更新記錄的代碼。我沒有看到任何問題，但是它沒有引發錯誤并指出它已成功運行。我嘗試了幾種其他方法，它們只是不更新MSSQL表中的行。數據庫/表在運行Windows 10和IIS的我的機器上是本地的（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>我到底在哪里錯。我可以連接，修改和創建就好了。我是否需要調整$ USER或查詢？
3 回答
慕姐4208626
TA貢獻1852條經驗 獲得超7個贊
我犯了一個菜鳥錯誤，我試圖在MSSQL中更新ID和時間戳列，這是不允許的。我將自己的代碼更改為此，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";
白板的微信
TA貢獻1883條經驗 獲得超3個贊
您是否直接在MSSQL Management Studio中嘗試了該查詢？
之后，您會看到它工作正常，請嘗試“回顯” POST變量以查看是否正確獲取了它們。
智慧大石
TA貢獻1946條經驗 獲得超3個贊
我沒看到INSERT。相反，您僅使用UPDATE意味著僅修改現有記錄。這是你想要的嗎？
$sql = "UPDATE users
SET id = :id,
firstname = :firstname,
lastname = :lastname,
email = :email,
age = :age,
location = :location,
date = :date
WHERE id = :id";
在這里，您通過選擇一個（所有）記錄id = :id，并在此處SET id = :id將其設置id為一個已設置的值。Is不會破壞某些內容，但這暗示您邏輯中的某些內容是錯誤的。
順便說一句，您允許每個客戶更改記錄。不是SQL注入，但很可能非常糟糕。
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