我在这里发布了我的问题,在我编辑帖子之前它已经关闭并不是一个真正的问题!
我有一个这样的登录表单:
<html>
<head>
<title>Password Checking Script</title>
</head>
<body>
<form action="check_user-pass.php" method="POST">
<h3>Please Login</h3>
User Name: <input type="text" name="user_name"><br>
Password: <input type="password" name="password">
<input type="submit" name="submit" value="Login!">
</form>
</body>
</html>
如您所见,此表单通过身份验证用户check_user-pass.php.
它在我的数据库中查找这些凭据; 如果它们存在,则返回OK,否则返回值NO.
所以我的问题是:我应该包含哪些代码check_user-pass.php？
我试图添加更多代码,但也不能这样做!我目前的代码是:
<?php
ob_start();
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password") or die(mysql_error());
echo "Connected to MySQL<br />";
mysql_select_db("$db_name") or die(mysql_error());
echo "Connected to Database<br />";
// Check $username and $password
/*
if(empty($_POST['username']))
{
echo "UserName/Password is empty!";
return false;
}
if(empty($_POST['password']))
{
echo "Password is empty!";
return false;
}
*/
// Define $username and $password
$username=$_POST['username'];
$password=md5($_POST['pass']);
// To protect MySQL injection (more detail about MySQL injection)
$username = stripslashes($username);
$password = stripslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);
$sql="SELECT * FROM $tbl_name WHERE username='$username' and password='$password'";
$result=mysql_query($sql);
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $username and $password, table row must be 1 row
if ($count==1) {
echo "Success! $count";
} else {
echo "Unsuccessful! $count";
}
ob_end_flush();
?>