Rob*_*eck 1 php mysql database
<?php
//Sessions starten
session_start();
//Include stuff
include('conf.php');
//'Shortcuts' maken
$sname = $site['name'];
$shost = $site['host'];
$shostb = $site['hostb'];
//Verbinding maken met DB
$link = mysql_connect($database['host'], $database['username'], $database['password']) or die (mysql_error());
mysql_select_db($database['db'],$link) or die(mysql_error());
//Gegevens uit db halen
if(isset($_GET['page'])){
$page['title'] = mysql_query("SELECT title FROM pages WHERE page = $_GET[page]");
}
else{
$page['title'] = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
}
?>
<!doctype html>
<html lang="nl">
<head>
<meta charset="utf-8" />
<title><?php echo $sname;?>•<?php echo $page['title']?></title>
有人知道为什么这不起作用？我的数据库结构是:
页面|标题
index.php文件|指数
对不起,但我不明白这一点.该变种转储的$page['title'] = resource(5)类型(MySQL的结果).我怎样才能解决这个问题？谢谢您的帮助.
mysql_query返回资源,您需要在资源上使用mysql_fetch_array(或类似mysql_fetch_assoc)获取标题值.
例如:
$res = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
$page = mysql_fetch_assoc($res);
// $page['title'] now contains the value (assuming there's an index.php page in the DB).
警告:您的查询容易受到SQL注入攻击.