我想爲我的網站做一個商店,所以我需要MySQL。 我想從用戶那裏拿走50金幣,並給予5點金幣。 我的當前腳本如下所示:需要MySQL數據庫的PHP腳本
<?php
ob_start();
session_start();
print('<meta http-equiv="content-type" content="text/html; charset=UTF-8"
/>');
include('config.php');
if(isset($_SESSION['rank']) && $_SESSION['rank'] >= 1)
{
if(isset($_SESSION['Gold']) && $_SESSION['Gold'] >= 50)
{
mysql_query("UPDATE users SET Gold = $_SESSION[Gold]-50 WHERE id =
$_SESSION[id];");
mysql_query("UPDATE users SET AtkDmg = $_SESSION[AtkDmg] + 5' WHERE id =
$_SESSION[id];");
} else header('location: shop.php');
} else header('location: login.php');
mysql_close()
?>
什麼是會話?這就是:
<?php
ob_start();
session_start();
include('config.php');
include('login_form.php');
if(isset($_POST["login"])){
$nickname = $_POST["nickname"];
$password = md5($_POST["password"]);
$lekerdezes = mysql_query("SELECT * FROM users WHERE nickname =
'".mysql_real_escape_string($nickname)."' AND password = '$password'");
$vanelekerdezes = mysql_num_rows($lekerdezes);
if ($vanelekerdezes>0)
{
header('location: login.php');
$adatok=mysql_fetch_assoc($lekerdezes);
$_SESSION["id"]=$adatok["id"];
$_SESSION['bann'] = 0;
$_SESSION["nickname"]=$adatok["nickname"];
$_SESSION["rank"]=$adatok["rank"];
$_SESSION["Gold"]=$adatok["Gold"];
$_SESSION["AtkDmg"]=$adatok["AtkDmg"];
}
else
{
print 'Hibás felhasználónév vagy jelszó!';
print mysql_error();
}
} else if(isset($_SESSION["nickname"])){
header('location: home.php');
}
?>
我希望你能幫助我,我還在學習PHP的,所以也許這就是爲什麼我不能修復這個簡單的事......所以,如果你這樣寫:學習PHP的,我「M已經這樣做:)
的sooo ...這是什麼問題?你有什麼錯誤嗎?也許可以快速瀏覽[如何提出問題](https://stackoverflow.com/help/how-to-ask)? – Naruto
它沒有做任何事情,我得到一個空白頁面,抱歉的問題不好。 – hunzeno
'mysql_ *'api已棄用嘗試使用'mysqli_ *或pdo' – JYoThI