Question

运行以下更改密码脚本时出现此错误：

注意：未定义的变量：第21行的C：\ xampp \ htdocs \ oh \ change.php中的$ insert 密码未更改

<?php
include('db_connection.php');
include('crypt.php');
session_start();
if (isset($_POST['submit'])) {
    $oldpassword = $_POST['current_password'];
    $newpassword= $_POST['new_password'];
    $confirm_password =$_POST['confirm_password'];
    $user_name = $_SESSION['UserName'];          
    $old = decrypt($oldpassword);
    $select = mysql_query("SELECT * FROM staff WHERE UserName='$user_name'");
    $fetch = mysql_fetch_array($select);
    $data_password = $fetch['password';
    if ($newpassword == $confirm_password && $data_password == $old) {
        $pass = encrypt($confirm_password);
        $insert = mysql_query("UPDATE staff SET password='$pass' WHERE UserName='$user_name'");
    }
    if ($insert) {
        echo "Password changed";
    } else {
        echo "Password  not changed";
    }
}
mysql_close($con);
?>

Answer 1

嗯，这是因为您可能无法进入if语句中设置$insert。如果是$insert不存在，则会收到此通知。

尝试检查变量是否存在，如：

if (isset($insert)) {
    echo "Password changed";
} else {
    echo "Password  not changed";
}

Answer 2

你应该在if语句之前定义$ insert，所以当你稍后引用时，肯定会定义它：

$insert = FALSE;
if ($newpassword == $confirm_password && $data_password == $old) {
    $pass = encrypt($confirm_password);
    $insert = mysql_query("UPDATE staff SET password='$pass' WHERE UserName='$user_name'");
}
if ($insert) {
    echo "Password changed";
} else {
    echo "Password  not changed";
}

编辑：

还有另一个问题，因为旧密码可能会在数据库中加密，并且您将其与提供的$ oldpassword的解密版本进行比较。尝试将该行更改为：

$old = encrypt($oldpassword);

Answer 3

正在发生的事情是$ insert未在之前的“if”语句之外声明。如果此语句为false，则$ insert未定义。试试这个：

    <?php
    include('db_connection.php');
    include('crypt.php');
    session_start();
    if (isset($_POST['submit'])) {
        $oldpassword = $_POST['current_password'];
        $newpassword= $_POST['new_password'];
        $confirm_password =$_POST['confirm_password'];
        $user_name = $_SESSION['UserName'];          
        $old = decrypt($oldpassword);
        $select = mysql_query("SELECT * FROM staff WHERE UserName='$user_name'");
        $fetch = mysql_fetch_array($select);
        $data_password = $fetch['password';
        if ($newpassword == $confirm_password && $data_password == $old) {
            $pass = encrypt($confirm_password);
            $insert = mysql_query("UPDATE staff SET password='$pass' WHERE UserName='$user_name'");

        if ($insert) {
            echo "Password changed";
        } else {
            echo "Password  not changed";
        }
}
    }
    mysql_close($con);
    ?>

＆＃34;注意：未定义的变量＆＃34;

3 个答案: