当让我登录时，它没有显示用户数据。

Question

我正在制作一个简单的登录脚本，当选中单选按钮时，应该让用户登录。如果选中“让我登录”单选按钮，我已经设置了一个24小时后到期的cookie，而且它有效.但是我发现了一些错误，比如: C:\xampp\htdocs\Webintegrator-PHP\MySql&PHP\Register&Login\core_login.php中的未定义索引: user_id，第17行，&警告: mysql_result()：无法跳到第20行上的MySQL结果索引6上的0行。

这些是用户的姓名和姓氏，应该从数据库中显示。

但是它们总是在用户登录时显示，直到单选按钮“让我登录”被选中为止。

以下是login_form.php的代码

<?php

if(isset($_POST['username']) && isset($_POST['password'])) {
    $username = $_POST['username'];
    $password = $_POST['password'];

    $password_hash = md5($password);

    if(!empty($username) && !empty($password)){
        $query = "SELECT id FROM users WHERE username = '" . mysql_real_escape_string($username) . "' AND password = '" . mysql_real_escape_string($password_hash) . "'";
        if($query_run = mysql_query($query)){
            $query_num_rows = mysql_num_rows($query_run);

            if($query_num_rows == 0) {
                echo "Invalid username/password combination";
            } elseif ($query_num_rows == 1) {

                if(isset($_POST['remember_me'])){
                    setcookie("remember_username", $_POST['username'], time()+3600*24);
                    setcookie("remember_password", $_POST['password'], time()+3600*24);

                    $user_id = mysql_result($query_run, 0, 'id');
                    $_SESSION['user_id'] = $user_id;

                    header ('Location: index_login.php');

                } elseif(isset($_POST['keep_me'])){
                    $user_id = mysql_result($query_run, 0, 'id');
                    setcookie('keep_me_loggedin', $user_id, time()+3600*24);

                    header ('Location: index_login.php');

                } else {
                    $user_id = mysql_result($query_run, 0, 'id');
                    $_SESSION['user_id'] = $user_id;

                    header ('Location: index_login.php');                    
                }
            }
        }
    } else {
        echo "You must supply a username & password";
    }
}

?>

<form action="<?php echo $current_file; ?>" method="POST">
    Username:<input type="text" name="username" value="<?php if(isset($_COOKIE['remember_username'])) {echo $_COOKIE['remember_username'];} ?>"><br/><br/>
    Password:<input type="password" name="password" value="<?php if(isset($_COOKIE['remember_password'])) {echo $_COOKIE['remember_password'];} ?>"><br/><br/>
    <input type="radio" name="remember_me"> Remember me?<br/><br/> 
    <input type="radio" name="keep_me"> Keep me logged in?<br/><br/> 
    <input type="submit" name="submit" value="Log in">
</form>

和index_login.php

<?php

require 'connect_login.php';
require 'core_login.php';

if(loggedin()) {
    $firstname = getuserfield('firstname');
    $surname = getuserfield('surname');
    echo "You are logged in, " . "$firstname " . "$surname " . "<a href='logout.php'>Log out</a>";
} else {
    include 'login_form.php';
}

?>

这是core_login.php

<?php

ob_start();
session_start();

$current_file = $_SERVER['SCRIPT_NAME'];

function loggedin(){
    if (isset($_SESSION['user_id']) && !empty($_SESSION['user_id']) || (isset($_COOKIE['keep_me_loggedin'])) && (!empty($_COOKIE['keep_me_loggedin']))) {
        return true;
    } else {
        return false;
    }
}

function getuserfield($field) {
    $query = "SELECT $field FROM users WHERE id = '" . $_SESSION['user_id'] . "'";

    if ($query_run = mysql_query($query)) {
        return mysql_result($query_run, 0, $field);
    }    
}


?>

任何解决问题的办法都将不胜感激！

提前感谢

Answer 1

似乎你只是需要做一些额外的条件检查。

检查$query_run是否是有效的mysql查询结果。

在尝试将isset放在getuserfield()中之前，要检查isset($_SESSION‘’user_id‘)是否为真。

编辑：

我在PHP ()页面上找到了这样的内容：

The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.

这意味着您应该使用mysql_fetch_array()而不是mysql_result()来访问执行的查询的结果。

结果()