php - 登录表单无输出

Question

这是登录 php 脚本，我没有收到任何错误或 echo ，所以我不知道发生了什么。 “*”是我审查该信息。如果您想查看实时页面，请访问 here。 .

<?php
printf("Client library version: %s\n", mysqli_get_client_info());

session_start();

$username = $_POST['username'];
$password = $_POST['password'];
var_dump('test');
if (!empty($username) && !empty($password)){
    echo $username
    echo $password

$connect = mysql_connect("localhost", "atsbusin_atsbusin", "Andrew85") or die ("Failed to connect to DB");
mysql_select_db("atsbusin_users") or die ("failed to select DB");

$query = mysql_query("SELECT * FROM users WHERE username='$username'");
$numrows = mysql_num_rows($query);
    if ($numrows != 0)
        {
            while ($row = mysql_fetch_assoc($query))
            {
                $dbusername = $row['username'];
                $dbpassword = $row['password'];
            }

                if ($username == $dbusername && $password == $dbpassword)
                {
                    echo "Login Success!";
                    $_SESSION['username'] == $dbusername;
                } else {
                    echo "Incorrect password!";
                }

        } else {
            die("invalid username, check the spelling or register that name");
        }

}
    echo 'something'
?>

这是最终的工作代码，感谢所有提供帮助的人!现在让我切换到 mysqli 并执行建议的安全措施(感觉我会回来☺)

<?php

session_start();

$username = $_POST['username'];
$password = $_POST['password'];

$connect = mysql_connect("localhost", "atsbusin_atsbusi", "Andrew85") or die ("Failed to connect to DB");
mysql_select_db("atsbusin_users") or die ("failed to select DB");

$query = mysql_query("SELECT * FROM users WHERE username='$username'");
$numrows = mysql_num_rows($query);
            while ($row = mysql_fetch_assoc($query))
            {
                $dbusername = $row['username'];
                $dbpassword = $row['password'];

            }

            if ($username == $dbusername && $password == $dbpassword)
            {
                echo "Login Success!";
                $_SESSION['username'] = $dbusername;
            } else {
                echo "Incorrect password!";
            }
?>

Answer 1

这不会回显任何内容并且不会给您错误的唯一方法是如果用户名存在且密码不匹配那么它将通过:

if($numrows != 0){

}

尝试在该 IF 语句中放置一个 var_dump('test'); 并查看它是否显示某些内容。

在

之后你也没有 ;

$username = $_POST['username']
$password = $_POST['password'] 

将其更改为:

$username = $_POST['username'];
$password = $_POST['password'];

一些额外的注意事项:

1. 您可能需要一个else:die("Do not leave username and/or password field blank");
2. 不要使用 mysql，它会从 PHP 5.5.0 开始弃用，并会在未来删除。
3. 您对 MYSQL 注入(inject)持开放态度

编辑:

如果您填写的用户名和密码不匹配，它也不会显示错误或回显