我正在尝试创建两个单独的会话 - 一个用于用户是管理员而另一个用户是作者。 $ type存储类型为枚举(可以是作者或管理员)。但是,我没有为我的会话获得唯一的会话ID。我是PHP和MySQL的新手。谁能告诉我我的代码中的错误在哪里。 请帮忙
<?php
session_start();
?>
<?php
include("dbconnect.php");
$con= new dbconnect();
$con->connect();
//create and issue the query
$sql = "SELECT type FROM users WHERE username = '".$_POST["username"]."' AND password = PASSWORD('".$_POST["password"]."')";
$result = mysql_query($sql);
//get the number of rows in the result set; should be 1 if a match
if (mysql_num_rows($result) == 1) {
$type_num=0;
//if authorized, get the values
while ($info = mysql_fetch_array($result)) {
$type =$info['type'];
}
if($type == "admin")
{
$_SESSION['type']=1;
$u = 'welcome.php';
header('Location: '.$u);
}
else
{
$_SESSION['type']=$type_num;
$u = 'welcome.php';
header('Location: '.$u);
}
}
else {
//redirect back to loginfailed.html form if not in the table
header("Location: loginfailed.html");
exit;
}
?>
welcome.php如下:
<?php
session_start();
?>
<html>
<body>
<h2>Welcome to SOD73.</h2>
<?
if($_SESSION['type']==1){
echo "You are of the usertype Admin and your session id is ";
echo session_id();
session_destroy();
}
else {
echo "You are of the usertype Author and your session id is ";
echo session_id();
session_destroy();
}
?>
</body>
</html>