我使用phpmyadmin和我的数据库有权限的所有特权。文本字段被命名为php代码,我没有得到任何错误从dreamweaver。当我在wamp localhost中测试时。我得到"无效的登录信息"…我在我的表"users"中创建了一个用户名和密码,并添加了Martin作为用户和123456作为密码…请检查我的代码?
<?php
if (isset($_POST['username']))
{
$username = $_POST['username']; //martin
$password = md5($_POST['password']); //123456
// connect to server
$con = mysql_connect("localhost", "root", "");
if(!$con){
die('Could not connect: '. mysql_error());}
mysql_select_db("test", $con);
if(mysql_num_rows(mysql_query("SELECT * FROM users where usermame = '$username' AND
password = '$password'")))
{//Correct information
$sql = "SELECT * FROM users where username = '$username' AND password = '$password'";
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
while($row = mysql_fetch_array($result))
{
$expire = time()+60*60*24*30;//1 month
setcookie("id", $row['id'], $expire);
echo "Logged in as <b>".$row['username']."</b>";
}
}else{
//false information
echo "Invalid login information.";
}
mysql_close($con);
echo $_COOKIE['id'];
}
?>
if(mysql_num_rows(mysql_query("SELECT * FROM users where usermame = '$username' AND
password = '$password'")))
您正在使用'username'代替上面代码中的'username'。拼写错误