Question

我很抱歉提出这么一个狭隘的问题，但我在PHP中有这个代码，它应该更新用户的帐户。没有返回错误，我的IDE也无法识别问题。现在的问题是代码没有更新数据库。我希望我能就这个问题得到一些帮助。

这是我的PHP代码：

<?php

    session_start();

    $con = mysqli_connect("mysql.serversfree.com", "u190182631_embo", "17011998embo", "u190182631_login");

    $username = $_POST['user_name']; 
    $last = $_POST['lname'];
    $first = $_POST['fname'];
    $address = $_POST['address'];
    $email = $_POST['email'];
    $year = $_POST['year'];

    if ($_SERVER["REQUEST_METHOD"] == "POST") {
        if (mysqli_connect_errno()) {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }

        mysqli_query($con,"UPDATE users SET last_name = '$last' 
        WHERE user_name = $_SESSION[user_name]");
        mysqli_close($con);
    }   
?>

如果需要，我的任何HTML表单：

<form method="post" action="update.php">
     Username: <input type="text" name="user_name" value="<?php echo $_SESSION['user_name']?>"><br><br>
     Email: <input type="text" name="email" value="<?php echo $_SESSION['user_email']?>"><br><br>
     Last Name: <input type="text" name="lname" value="<?php echo $_SESSION['last_name']?>"><br><br>
     First Name: <input type="text" name="fname" value="<?php echo $_SESSION['first_name']?>"><br><br>
     Street Address: <input type="text" name="address" value="<?php echo $_SESSION['address']?>"><br><br>
     Year Graduated: <input type="text" name="year" value="<?php echo $_SESSION['year']?>"><br><br>
     <input type="submit" value="Update Information"><br>
 </form>
 <form method="link" action="manage.php">
         <input type = "submit" value = "Cancel"><br> 
 </form>

任何帮助都会很棒！

Answer 1

试试这个 - 它也有助于防止SQL注入攻击：

$db = new mysqli("mysql.serversfree.com", "u190182631_embo", "17011998embo", "u190182631_login");

$username = $_POST['user_name']; 
$last = $_POST['lname'];
$first = $_POST['fname'];
$address = $_POST['address'];
$email = $_POST['email'];
$year = $_POST['year'];


if ($_SERVER["REQUEST_METHOD"] == "POST")
{
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $stmt = $db->prepare("UPDATE users SET last_name = ? AND WHERE user_name = ?;");
    $stmt->bind_param("ss", $last, $_SESSION['user_name']);
    $stmt->execute();
    $stmt->close();
}

Answer 2

这里的一个大问题是你不知道如何自己调试问题，也不知道在请求帮助中包含哪些信息。

没有返回错误

你怎么知道的？您不检查查询中的任何错误。考虑：

$upd="UPDATE users SET last_name = '$last' 
    WHERE user_name = $_SESSION[user_name]";
if (!mysqli_query($con,$upd)) {
   print "query failed: $upd \n\n<br />" . mysqli_error();
}

您已经展示了用于生成表单的代码片段 - 但不是实际发送到浏览器的代码，

正如Fred -ii-所说，你的SQL中没有引用$ _SESSION [user_name]似乎很奇怪。

Answer 3

试试这个

mysqli_query($con,"UPDATE users SET last_name = '$last' WHERE user_name = {$_SESSION['user_name']}");

Answer 4

更新此行代码：

mysqli_query($con,"UPDATE users SET last_name = '$last' 
WHERE user_name = $_SESSION[user_name]");

使用新的：

mysqli_query($con,"UPDATE users SET last_name = '$last' 
WHERE user_name = $_SESSION['user_name']");

希望它能奏效！

PHP代码不更新数据库，没有sql错误，我的IDE没有返回错误，但数据库没有更新

4 个答案: