php用户cookie登录验证与mysql数据登录验证的方法:
<?php
unset($username);
if ($_COOKIE['login']) {
list($c_username,$cookie_hash) = split(',',$_COOKIE['login']);
if (md5($c_username.$secret_word) == $cookie_hash) {
$username = $c_username;
} else {
print "You have sent a bad cookie.";
}
}
if ($username) {
print "Welcome, $username.";
} else {
print "Welcome, anonymous user.";
}
看个完整的与数据库连接登录的代码:
<html> <head> <title>Log-In Page</title> </head> <body> Please enter your user details to log-in here... <form action = "authenticate.php" method = "post"> Username:<br> <input type = "text" name = "username"> <br><br> Password:<br> <input type = "text" name = "password"> <br><br> <input type = "submit" value = "Log In"> </form> </body> </html>
File:authenticate.php
class="brush:php;"><?php
$username = $_POST['username'];
$password = $_POST['password'];
$self = $_SERVER['PHP_SELF'];
$referer = $_SERVER['HTTP_REFERER'];
if( ( !$username ) or ( !$password ) )
{ header( "Location:$referer" ); exit(); }
$conn=@mysql_connect( "localhost", "userName", "password" ) or die( "Could not connect" );
$rs = @mysql_select_db( "my_database", $conn ) or die( "Could not select database" );
$sql = "select * from users where user_name="$username" and password = password( "$password" )";
$rs = mysql_query( $sql, $conn ) or die( "Could not execute query" );
$num = mysql_numrows( $rs );
if( $num != 0 )
{
$msg = "<h3>Welcome $username - your log-in succeeded!</h3>";
}
else
{
header( "Location:$referer" ); exit();
}
<html>
<head>
<title>Log-In Authenticated</title>
</head>
<body>
<?php echo( $msg );
</body>
</html>
jQuery响应式后台登录界面模板html源码,登录页面通过jquery来验证表单,判断用户名和密码是否符合要求,通常登录页面在企业网站或者商城网站都是必须要用到的页面,响应式的后台页面,当浏览器放大或者缩小,背景会根据浏览器来调整图片的大小!php中文网推荐下载!
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