Question

我正在使用phpmyadmin，我的数据库拥有所有权限的权限。 Textfields在php代码中命名，我没有从Dreamweaver得到任何错误。当我在wamp localhost中测试它时。我收到“无效登录信息”...我在我的表中创建了用户名和密码“users”还添加了martin作为用户，123456作为密码...请查看我的代码？

<?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'];
}
?>

Answer 1

 if(mysql_num_rows(mysql_query("SELECT * FROM users where usermame = '$username' AND 

    password = '$password'")))

您在上面的code.spelling错误

中使用'usermame'代替'username'

PHP - 登录表单。我一直收到“无效的登录信息”

1 个答案: