<?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 $page['title']?></title>有人知道为什么这不管用吗?我的数据库结构是:
页面标题
index.php|index
对不起我搞不清楚。var 转储类型的$page['title'] = resource(5) (mysql结果)。我怎么才能解决这个问题?谢谢你的帮助。
发布于 2012-03-20 21:41:48
mysql_query返回一个资源,您需要在资源上使用mysql_fetch_array (或类似的,例如mysql_fetch_assoc)来获取标题值。
例如:
$res = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
$page = mysql_fetch_assoc($res);
// $page['title'] now contains the value (assuming there's an index.php page in the DB).警告:查询容易受到SQL注入的影响。
发布于 2012-03-20 21:45:09
这是由mysql_query返回的resource,在资源上使用mysql_fetch_assoc获取标题值。
做以下修改:
if(isset($_GET['page'])){
$result = mysql_query("SELECT title FROM pages WHERE page = $_GET[page]");
}
else{
$result = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
}
$row= mysql_fetch_assoc($result);
$page['title'] =$row['title'];发布于 2012-03-20 21:45:27
您需要使用像mysql_fetch_assoc()或mysql_result()这样的函数来获取实际数据。
if(isset($_GET['page'])){
$result = mysql_query("SELECT title FROM pages WHERE page = '" . mysql_real_escape_string($_GET[page]) . "'");
$page['title'] = mysql_result($result, 0, 'title');
}
else{
$result = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
$page['title'] = mysql_result($result, 0, 'title');
}https://stackoverflow.com/questions/9795422
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