php $stmt返回 0？-解网

问：

这是我的代码：

if (isset($_GET['email'], $_GET['code'])) {
    if ($stmt = $con->prepare('SELECT * FROM accounts WHERE email = ? AND activation_code = ?')) {
        $stmt->bind_param('ss', $_GET['email'], $_GET['code']);
        $stmt->execute();
        // Store the result so we can check if the account exists in the database.
        $stmt->store_result();
        if ($stmt->num_rows > 0) {
            // Account exists with the requested email and code.
            if ($stmt = $con->prepare('UPDATE accounts SET activation_code = ? WHERE email = ? AND activation_code = ?')) {
                // Set the new activation code to 'activated', this is how we can check if the user has activated their account.
                $newcode = 'activated';
                $stmt->bind_param('sss', $newcode, $_GET['email'], $_GET['code']);
                $stmt->execute();
                echo 'Your account is now activated! You can now <a href="index.html">login</a>!';
            }
        } else {
            echo 'The account is already activated or doesn\'t exist!';
        }
    }
}

我总是得到这个输出：

echo 'The account is already activated or doesn\'t exist!';

我在这里做错了什么？

php mysql mysqli

if (isset($_GET['email'], $_GET['code'])) {
    if ($stmt = $con->prepare('SELECT activation_code FROM accounts WHERE email = ?')) {
        $stmt->bind_param('s', $_GET['email']);
        $stmt->execute();
        // Store the result so we can check if the account exists in the database.
        $stmt->store_result();
        if ($stmt->num_rows > 0) {
            // Account exists with the requested email
            $result = $stmt->get_result();
            $row = $result->fetch_assoc();
            if ($row['activation_code'] != $_GET['code']) {
                if ($stmt = $con->prepare('UPDATE accounts SET activation_code = ? WHERE email = ?')) {
                    // Set the new activation code to 'activated', this is how we can check if the user has activated their account.
                    $newcode = 'activated';
                    $stmt->bind_param('ss', $newcode, $_GET['email']);
                    $stmt->execute();
                    echo 'Your account is now activated! You can now <a href="index.html">login</a>!';
                }
            } else {
                echo 'The account is already activated.';
            }
        } else {
            echo 'The account doesn\'t exist!';
        }
    }
}

我收到此错误：致命错误：未捕获mysqli_sql_exception：命令不同步;您现在无法在 /home/u590333097/domains/mysite.com/public_html/activate.php：21 中运行此命令堆栈跟踪： #0 /home/u590333097/domains/mysite.com/public_html/activate.php（21）： mysqli_stmt->get_result（） #1 {main} 扔在 /home/u590333097/domains/mysite.com/public_html/activate.php 上 21 行

0赞 Barmar 9/2/2023

有没有可能有多行相同？仅当您尚未读取第一个查询中的所有行时，才会发生这种情况。email

0赞 Barmar 9/2/2023

请参阅 stackoverflow.com/questions/3632075/...

0赞 Ben5678 9/2/2023

只有 1 行电子邮件

1赞 Barmar 9/2/2023

然后我不确定为什么会这样。也应该防止这种情况发生。当您访问此代码时，您是否还有其他疑问？$stmt->store_result()

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