我想获取页面中除用户名和用户 ID 之外的信息。关于这一点，我创建了两个 php 页面。我的数据库也由 3 列 userid、用户名、密码组成。 login.php 页面是
<?php
session_start();
//@$userid = $_GET['userid'];
@$username = $_POST['username'];
@$password = $_POST['pass'];
if(@$_POST['Submit']){
if($username&&$password)
{
$connect = mysql_connect("localhost","*****","") or die("Cannot Connect");
mysql_select_db("project") or die("Cannot find the database");
$query = mysql_query("SELECT * FROM users WHERE username='$username'");
//$query = mysql_query("SELECT * FROM users WHERE userid='$userid' and username='$username'");
$numrows = mysql_num_rows($query);
if($numrows!=0)
{
while ($row = mysql_fetch_assoc($query))
//while ($row = mysql_fetch_array($query))
{
$dbuserid = $row['userid'];
$dbusername = $row['username'];
$dbpassword = $row['password'];
}
if($username==$dbusername&&$password==$dbpassword)
{
echo "You are login!!!!! Continue now with the survey <a href='mainpage.php'>here</a>";
$_SESSION['username']=$username;
$_SESSION['userid']=$userid;
}
else
{
echo "<b>Incorrect Password!!!!</b>";
}
}
else
//die("That user does not exist");
echo "<b>That user does not exist</b>";
}
else
echo "<b>You must enter a username and a password</b>";
}
?>
<!--<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">-->
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<!--<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />-->
<title>Login Page</title>
<style type="text/css">
h2 {letter-spacing: 10px; font-size: .2in; background-color: #33CC00; color: #000000; text-transform:uppercase; width:260px}
span {color: #FF00CC}
legend {font-variant: small-caps; font-weight: bold}
fieldset {width: 260px; height: 100px; font-family: "Times New Roman", Times, serif; background-color: #CCCCCC; color: #000000}
label {display:block;}
.placeButtons {position: relative; left: 0px; width: 70px; margin: 5px; 0px;}
</style>
</head>
<body background="images/good.jpg">
<h2>Login Page</h2>
<form name="loginform" method='POST'>
<fieldset>
<legend>Form</legend>
<label>Username: <input type="text" name="username"/><span>*</span></label><br/>
<label>Password: <input type="password" name="pass"/><span>*</span></label>
<input class="placeButtons" type="reset" value='Reset'/>
<input class="placeButtons" type="submit" name="Submit" value='Login'/>
<a href='registration.php'>Register</a>
</fieldset><br>
<a href='firstpage.php'><-- Go Back</a>
</form>
</body>
</html>
以及用户的欢迎页面
<?php
session_start();
if ($_SESSION['username'])
{
//echo "Welcome, ".$_SESSION['username']."! <a href='logout.php'>Logout</a>";
echo "Welcome, ".$_SESSION['username']."<br>".$_SESSION['userid']. "<a href='logout.php'>Logout</a>";
}
else
die("You must be logged in!!");
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<!--<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />-->
<title></title>
</head>
<body background="images/good.jpg">
</body>
</html>
问题是在欢迎页面中它只向我显示用户名而不是用户 ID。我缺少什么？此外，我知道我的登录页面不是最好的，是SQL注入(inject)攻击的典型例子。我必须改进它。
最佳答案
我注意到了一件很快的事。这可能就是问题所在。 $_SESSION['userid'] 正在从未设置的 $userid 获取值。另外，使用 @ 来抑制错误也不是一个好的做法。使用 isset 检查变量是否已设置并继续。
$_SESSION['userid'] = $userid; //where are you getting $userid from?
这应该是
$_SESSION['userid'] = $dbuserid;
也不要使用类似的语句
if ($_SESSION['username'])
首先检查变量是否像这样设置
if ( isset($_SESSION['username']) ){
//now continue your work
}
关于php - 如何在 session 中显示用户 ID？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18162821/