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我想知道是否有人可以帮助我，我目前正在做一个登录，将用户引导到页面，以便他们可以玩游戏。但是因为在发送信息时数据库上有多个帐户，它会为数据库上的所有帐户而不是一个用户显示它，任何关于如何让它将信息提交给其中一个用户的想法，如果这样做对你有意义？？

这是我的登录部分，希望对您有所帮助：

<?php

include('connecttest.php');

session_start(); //we're using sessions so this is required!

if($_SESSION['loggedin'] == TRUE) {
header('location: http://html5b.mytestbox.co.uk/sessions/VolumeGame/VolumeGame1.html'); //members area
}else{

if($_POST['submitLogin']) { 
//verify login from user input

$username = mysql_real_escape_string($_POST['username']);
$password = md5(mysql_real_escape_string($_POST['password']));

$select_user = mysql_query("SELECT COUNT(id) AS amount FROM users WHERE username = '$username' AND password = '$password' ");
$user = mysql_fetch_assoc($select_user);
$amount_found = (int)$user['amount']; //amount of users found by the query

if($amount_found > 0) {
$login_attempt = 1; //successful login attempt
$_SESSION['loggedin'] = TRUE;
$_SESSION['username'] = $username;
header('location: http://html5b.mytestbox.co.uk/sessions/VolumeGame/VolumeGame1.html'); //members area
}else{
$login_attempt = 0; //invalid login attempt
}

}

if( ($_POST['submitLogin'] AND isset($login_attempt) AND $login_attempt = 0) OR !$_POST['submitLogin'] ) {

//show login form

if($_POST['submitLogin']) { //attempted to login? (-> invalid login)
echo "<p>Invalid login. </p>";
}

?>

<form method="POST" action="login.php">
<b>Username:</b> <br /> <input type="text" name="username"> <p>
<b>Password:</b> <br /> <input type="password" name="password"> <p>
<input type="submit" name="submitLogin" value="Login!">
</form>

<?php

}

}

?>

这是将信息发送到数据库的代码：

    <?php
    include ("connecttest.php");

    session_start();

    $username = ($_POST['username']);

    $_SESSION['username'] = $username;

    $q1btn=$_POST['q1btnPressed'];
    $q2btn=$_POST['q2btnPressed'];
    $q3btn=$_POST['q3btnPressed'];
    $q4btn=$_POST['q4btnPressed'];


    if ($_POST['q1btnPressed'])
    {
        $sql=" UPDATE users
          SET q1answer = '$q1btn' 
          WHERE username = '$username'" ;
    }
    if ($_POST['q2btnPressed'])
    {
        $sql=" UPDATE users ;
          SET q2answer = '$q2btn'" ;
    }
    if ($_POST['q3btnPressed'])
    {
        $sql=" UPDATE users ;
          SET q3answer = '$q3btn'" ;
    }

          mysql_query($sql)or die (mysql_error);


    mysql_close();  
    ?>

<?php
include "connecttest.php";


$result = mysql_query("SELECT *  FROM users");

$rows=mysql_fetch_row($result);
    echo json_encode($rows);


?>
4

1 回答 1

0

向您的查询添加一个WHERE包含唯一用户 ID 的子句。

于 2013-01-14T11:57:55.260 回答