登录页面不直接-显示空白页面PHP
[英]Login Page doesnt direct - shows a blank page PHP
问题描述
所以我有一个登录脚本，不太确定为什么它不起作用。 在我的用户表中，我有以下字段：
表：用户
Field 1) ID
Field 2) Password (it is stored with crypt)
Field 3) Status (ranges from 0-2)
的index.php
<?php
session_start();
unset($_SESSION['Status']);
?>
<head>
<title>Login Form</title>
<link rel="stylesheet" type="text/css" href="login.css">
<img src="logo1.jpg" style="float:left; width:490px; height:130px; margin-top: -70px;">
<!--[if lt IE 9]><script src="//html5shim.googlecode.com/svn/trunk/html5.js"></script><![endif]-->
</head>
<body>
<section class="container">
<div class="login">
<h1>Login</h1>
<form action="process_login.php" method="POST"/>
<div class="help-tip">
<p>Enter the User ID and Password that you were given, in order to login. If you have forgotten your ID or Password, contact Admin</p>
</div>
<p><input type="number" name="ID" value="" placeholder="ID*" required autofocus></p>
<p><input type="password" name="Password" value="" placeholder="Password*" required></p>
<p class="submit"><input type="submit" name="commit" value="Login"></p>
</form>
</div>
process_login.php
<?php
session_start();
?>
<?php
//Connect to host site and databse
include("functions.php");
// Fetching variables
$id = $_POST['ID'];
$pw = crypt($_POST['Password']);
//Find user details from User table using the username entered and comparing the entered password with the one retrieved form the user table
$UserValidate = mysqli_query ("SELECT * FROM User WHERE ID = '$id'") or die (mysqli_error());
$row = mysqli_fetch_array($UserValidate);
$CorrectId = $row['ID'];
$CorrectPw = $row['Password'];
$UserType = $row['Status'];
//check if ID in database
if ($id == $CorrectId) {
//check if password is assigned to that username and is correct
if ($pw == $CorrectPw) {
//check if user is standard user
if ($UserType == 0) {
$_SESSION['CadetUser'] = $id;
header('http://****/calendar.php:'.$url);die();
if ($UserType == 1) {
$_SESSION['StaffUser'] = $id;
header('http://****/calendar_staff.php:'.$url);die();
if ($UserType == 2) {
$_SESSION['AdminUser'] = $id;
header('http://****/calendar_admin.php:'.$url);die();
}
}
else {
echo "Either your ID or Password is wrong";
header('http://******/index.php:'.$url);die();
}
}
}
}
?>
更新我的问题是，使用正确的详细信息登录时，出现黑屏。 它只是在process_login.php处停止。我也将重定向更改为“ header ..........”，如建议的那样
4 个解决方案
解决方案1
1 2016-04-13 17:16:31
对于重定向，您可以尝试
header('location:'.$url);die();
注意：在标头之前删除所有回显或打印，并确保在php开头标签之前没有空格
解决方案2
1 2016-04-13 18:14:02
顺便说一句，您的SQL语句容易受到SQL注入的攻击，因为您直接将$ id放入了语句中。 使用参数和mysqli会更安全
解决方案3
0 已采纳 2016-04-13 18:05:09
这是您的代码-缩进，如您在帖子中输入的那样：
//check if ID in database
if ($id == $CorrectId) {
//check if password is assigned to that username and is correct
if ($pw == $CorrectPw) {
//check if user is standard user
if ($UserType == 0) {
$_SESSION['CadetUser'] = $id;
header('http://****/calendar.php:'.$url);
die();
if ($UserType == 1) {
$_SESSION['StaffUser'] = $id;
header('http://****/calendar_staff.php:'.$url);
die();
// <----- missing a closing brace
if ($UserType == 2) {
$_SESSION['AdminUser'] = $id;
header('http://****/calendar_admin.php:'.$url);
die();
}
}
else {
echo "Either your ID or Password is wrong"; // you need to remove this; outputting HTML prior to sending headers will result in a PHP error
header('http://******/index.php:'.$url);
die();
}
}
}
} // <----- remove this
如您所见，唯一有机会的条件是if ($UserType == 0) 。 更不用说其中有一个错误的} ，它可能导致语法错误。
您的标题中也缺少Location ，例如。 header('Location: url/goes/here.php');
我在下面重新格式化了您的代码，并修复了语法错误：
//check if ID in database
if ($id == $CorrectId && $pw == $CorrectPw) {
//check if user is standard user
if ($UserType == 0) {
$_SESSION['CadetUser'] = $id;
header('Location: http://****/calendar.php:'.$url);
die();
}
elseif ($UserType == 1) {
$_SESSION['StaffUser'] = $id;
header('Location: http://****/calendar_staff.php:'.$url);
die();
}
elseif ($UserType == 2) {
$_SESSION['AdminUser'] = $id;
header('Location: http://****/calendar_admin.php:'.$url);
die();
}
else {
header('Location: http://******/index.php:'.$url);
die();
}
}
并且由于if ($id == $CorrectId)和if ($pw == $CorrectPw)是必须满足的条件才能继续进行，因此将它们包括在单个条件中是有意义的。 您应尽可能避免嵌套条件太深。 使事情变得凌乱，并且代码难以阅读/遵循。 您可以看到我已将它们添加到单个条件中。
解决方案4
0 2016-04-13 18:11:13
更改标题功能
header('http://****/calendar.php:'.$url);
至
header('location : http://****/calendar.php:');
在标题中添加位置，如上所示
登录页面显示空白
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