显示用户配置文件信息


Displaying users profile information

当点击他们的个人资料时,我有显示出用户个人资料信息的问题。所显示的数据是我自己登录时,所以它没有切换。我找不到问题,想知道我在哪里可能出错?

信息应该显示在profile。php

if (isset($_GET['username']) === true && empty ($_GET['username']) === false) {
    $username = $_GET['username'];
    if (user_exists($username) === true) {
        $user_id  = user_id_from_username($username);   
        $profile_data = user_data($user_id, 'first_name', 'last_name', 'email');        
    ?>  
    <h1><?php echo $profile_data['first_name']; ?> profile</h1>
    <p><?php echo $profile_data['email'] ?></p> 
    <?php
    } else {
        echo 'Sorry, that user does not exist';
    }
    } else {
        header('Location: index.php');
    exit();
}

显示的是我的信息,而不是我要查看的用户。如果我在URL上输入一个虚假的用户名,它会报错,说他们不存在。

下面是我从数据库中挑选数据的脚本:

if (logged_in() === true) {
    $session_user_id = $_SESSION ['user_id'];
    $user_data = user_data($session_user_id, 'user_id', 'username', 'password', 'first_name', 'last_name', 'email', 'type', 'profile');
    if(user_active($user_data['username']) === false) {
        session_destroy();
        header('Location: index.php');
        exit();
    }
}
$errors = array();

我使用。htaccess文件来初始化虚荣URL

RewriteEngine On
RewriteCon %{REQUEST_FILENAME} !-f
RewriteCon %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ /profile.php?username=$1

所以我的URL是这样的…

http://mywebsite.com/myname

下面是user_data 的函数
function user_data($user_id) {
    $data = array();
    $user_id = (int)$unser_id;
    $func_num_args = func_num_args();
    $func_get_args = func_get_args();
    if ($func_num_args > 1){
        unset($func_get_args[0]);
        $fields = '`' . implode('`, `', $func_get_args) . '`';
        $data = mysql_fetch_assoc(mysql_query("SELECT $fields FROM `users` WHERE 'user_id' = $user_id"));

        return $data;
    }
}

它通过更改myname而不是将个人资料页面上的用户数据切换到其他用户数据来工作。

user_exists() function

function user_exists($username) {
    $username = sanitize($username);
    $query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'");
    return (mysql_result($query, 0) == 1) ? true : false;
}

试试这个

function user_data($user_id, $fields = '*') {
  // type cast to int
  $user_id = (int) $user_id;
  // if an array is passed, we implode the array to get fields,
  // otherwise we return all rows
  if (is_array($fields))
  {
    $fields = implode('`, `', $fields) . '`';
  }
  //build query, with LIMIT 1 , I assume username is unique
  $qry = "SELECT {$fields} FROM `users` WHERE 'user_id' = {$user_id} LIMIT 1";
  $sql = mysql_query($qry);
  $result = mysql_fetch_assoc($sql);
  // if we have a result, return it, otherwise return false
  return (mysql_num_rows($sql) == 1) ? $result : false ;
}

使用:

$profile_data = user_data($user_id, array('user_id', 'username', 'password', 'first_name', 'last_name', 'email', 'type', 'profile'));

$fields = array(
  'user_id',
  'username',
  'password',
  'first_name',
  'last_name',
  'email',
  'type',
  'profile'
);
$profile_data = user_data($user_id, $fields);