php - undefined variable : Name in mysql PHP

Question

enter image description here如果登录信息错误，系统将正确显示“登录失败”。

但如果详细信息正确，系统将在第25行显示“C:\Users\Alex\Desktop\OSMAD\8.5\root\Project\test3.php”< br/>登录失败”

这里是 PHP 文本

<?php

    $host = "127.0.0.1";

    $username = "root";

    $password = "usbw";   

    $dbname =  "forum_system";

    // Connect to server
    $connect = mysql_connect($host, $username, $password) 
        or die ("Sorry, unable to connect database server");

    $dbselect = mysql_select_db($dbname,$connect) 
        or die ("Sorry, unable to connect database");

    $Name   = $_POST['Name'];
    $Password = $_POST['Password'];

    $result = mysql_query("select * from users where name = '$Name' and 
        password = '$Password'")
    or die("Sorry, query failed".mysql_error());

    $row = mysql_fetch_array($result);

    if ( $row['Name'] == $Name && $row['Password'] == $Password ) {
        echo "welcome";
    } else {
        echo "Fail login";
    }
?>

Answer 1

MySQL 表列名称区分大小写。因此，您应该使用 $row['name'] 和 $row['Name'] 和 $row['Password'] 来代替$row['密码']。所以这行:

if ( $row['Name'] == $Name && $row['Password'] == $Password ) 

应替换为:

if ( $row['name'] == $Name && $row['password'] == $Password )

也可以使用mysqli或MySQL_PDO而不是 mysql 函数。 Php 7.0.0 中删除了 mysql 扩展。