Php代码获得一个空白页面

Kat*_*lyn 0 php debugging login

此代码在浏览器中返回一个空白页面.你能发现任何错误吗?权限和网址已检查.phpcodechecker.com说它没关系

<?php
    # Check for POST login data, else set initial values
    if (isset($_POST["user"])) {
        $user=$_POST['user'];
        $pass=hash('sha256',$_POST['pass']);
    }
    else {
        $user="";
        $pass="";
    }

    # Check Login Data
    #
    # Password is hashed (SHA256). In this case it is 'admin'.
    if($user == "admin"
    && $pass == "8c6976e5b5410415bde908bd4dee15dfb167a9c873fc4bb8a81f6f2ab448a918")
    {
        # Load content from local storage
        include("../localstorage/content.html");
    }
    else
    {
        # Show login form. Request for username and password
echo
            '<html>
            <body>      
                <form method="POST" action="">
                    Username: <input type="text" name="user"><br/>
                    Password: <input type="password" name="pass"><br/>
                    <input type="submit" name="submit" value="Login">
                </form>
            </body>
            </html>'

    }
?>
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(我从http://www.canbike.org/information-technology/2014/02/05/php-simple-password-protection-with-session-timeout.html下载了它,我很清楚它缺乏安全性)

Gau*_*tam 6

使用echo语句打印html代码如下.

echo '<html>
            <body>      
                <form method="POST" action="">
                    Username: <input type="text" name="user"><br/>
                    Password: <input type="password" name="pass"><br/>
                    <input type="submit" name="submit" value="Login">
                </form>
            </body>
            </html>'; 
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并删除PHP标签之间,在开始和结束时只使用外部的PHP标签.