链接登录表单与数据库

-3

可能重复：
PHP/MySQL Login issue 链接登录表单与数据库

当试图登录至我的会员在网站上区域，它不链接与数据库，并因此进入零数据仍然带我到会员区。

下面的php和HTML有什么问题。

<?php 
session_start(); 
include("dbconnect.php"); 
$numrows=0; 
$sqlMember = mysql_real_escape_string($member); 
$sqlPassword = mysql_real_escape_string($password); 
$query="select First_name fname, Last_name lname, Email email, Membership_Number from members where (Membership_Number='$sqlMember' and Password='$sqlPassword')"; 

    $link = mysql_query($query); 
if (!$link) { 
    die('login error'); 
} 
$numrows=mysql_num_rows($link); 
if ($numrows>0){ // authentication is successfull 
    while($row = mysql_fetch_array($query)); 
    $_SESSION['user']['first_name']=$row['first_name']; 
    $_SESSION['user']['last_name']=$row['last_name']; 
    $_SESSION['user']['Membership_Number ']=$row['Membership_Number ']; 
    echo $_SESSION['user']['fname'].' '.$_SESSION['user']['lname']; 
    header("location:members_area.php"); 
} else { 
    header("location:../invalid.php"); // authentication was unsuccessfull 
} 
?>

HTML

<form id="jjjj" action="Send_log_details.php" class="register"> 
    <ul> 
    <li> 
    Membership Number:<br> 
    <input type="text" id="Membership_Number" name="Membership_Number"> 
    </li> 
     <li> 
    Password:<br> 
<input type="password" id="Password" name="Password" value="Log in"> 
     </li> 
    <li> 

    <input type="submit" name="register" value="Login" onclick="logMeIn()">    
     </li> 
     </ul> 
     </form>

来源

2013-01-11 Richard Jess

你还没有加入form.Then POST方法，你会如何获得$成员和$密码值登录。 – Arunu

$sqlMember = mysql_real_escape_string($member); 
$sqlPassword = mysql_real_escape_string($password);

应该

$sqlMember = mysql_real_escape_string($_POST["Membership_Number"]); 
$sqlPassword = mysql_real_escape_string($_POST["Password"]);

而且

<form id="jjjj" action="Send_log_details.php" class="register">

应该是

<form id="jjjj" action="Send_log_details.php" class="register" method="POST">

来源

2013-01-11 16:34:21

你的SQL对我来说似乎不对，我希望我不是。

// Your code 
$query="select First_name fname, Last_name lname, Email email, Membership_Number from members where (Membership_Number='$sqlMember' and Password='$sqlPassword')"; 

// My code 
$query="select fname AS First_name, lname AS Last_name, email AS Email, Membership_Number from members where (Membership_Number='$sqlMember' and Password='$sqlPassword')";

，此外，我只接受作为成功的，如果mysql_num_rows($query) == 1

来源

2013-01-11 16:39:28 Zim84

链接登录表单与数据库

回答

相关问题