【发布时间】:2011-10-13 12:21:10
【问题描述】:
我在 index.php 中有这个
<?php
include_once '..\connect.php';
session_start();
if (isset($_SESSION['username'])){
$player_name = $_SESSION['username'];
} else {
header( 'Location: http://localhost/Inventory/index.php' ) ;
exit;
}
?>
我正在向 request.php 发出 ajax 请求
<?php
//connect to databate and check for errors
$con = mysql_connect ("localhost","root","");
if (!$con) {
die ('Could not connect to database: ' . mysql_error());
}
//select database and check selection
if (!mysql_select_db ("GotA", $con)) {
die ('Could not select database: ' . mysql_error());
}
//I have to create this if not it doesnt find the sessions $player_name variable
$player_name = $_POST['name'];
//***Create Player Array**//
$player_info = "SELECT * from players where id = $player_name";
$player_info2 = mysql_query($player_info) or die ('Couldnt get players name');
$player_info3 = mysql_fetch_array($player_info2);
好吧,使用 javascript 发送的变量从数据库中检索数据似乎不安全,难道没有办法直接使用 index.php(会话部分)中的变量吗? 还是仅使用 javascript 传递信息是否安全?
【问题讨论】:
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请仔细阅读this question 及其接受的答案。明白它。使用它。
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谢谢你们所有的 cmets 和答案,我将通读它们,当我找到解决方案时链接会报告错误