我是PHP新手,試圖編寫一個登錄函數。位卡住並出現錯誤。 這裏我的功能:編寫php函數
<?php
if(!defined('_VALID_ACCESS')) die('direct access is not allowed.');
include('includes/connect.php');
function login($username, $password)
{
$username = trim($username);
$password = trim($password);
echo $username;
echo $password;
$login_sql = "SELECT * FROM user WHERE user = '".($username)."'
AND pass = '".(md5($password))."'";
$login_result = $mysqli->query($login_sql) or die(mysqli_error());
$row=$login_result->fetch_row();
if($row[0] == 1)
{
return true;
}
else
{
return false;
}
}
?>
connect.php
<?php
$db_name = "coolmates";
$db_server = "localhost";
$db_user = "justron";
$db_pass = "Justron9004";
$mysqli = new MySQLi($db_server, $db_user, $db_pass, $db_name) or die(mysqli_error());
?>
注意:未定義的變量:在mysqli的.. \ login.php中上線14
致命錯誤:調用一個成員函數查詢()在第14行上的.. \ login.php中的非對象上
幫助我。
更進一步,將登錄功能的末尾更改爲: return($ row [0] == 1); – 2009-05-02 05:32:42