资源ID#5:MySQL [英] Resource id #5: MySQL
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问题描述
<?php
//会话启动
session_start();
//包含内容
include(’conf.php’);
//快捷方式 maken
$ sname = $ site [’name’];
$ shost = $ site [’host’];
$ shostb = $ site [’hostb’];
//绑定数据库
$ link = mysql_connect($ database ['host'],$ database ['username'],$ database ['password'])或死掉(mysql_error()) ;
mysql_select_db($ database [’db’],$ link)或die(mysql_error());
// Gegevens uit db halen
if(isset($ set _ $ ['page'])){
$ page ['title'] = mysql_query(从页面中选择标题,而页面= $ _GET [page]);
}
else {
$ page [’title’] = mysql_query(从页面中选择标题,而页面='index.php');
}
?>
<!doctype html>
< html lang = nl>
< head>
< meta charset = utf-8 />
< title><?php echo $ sname;?>•<?php echo $ page [’title']?>< / title>
有人知道为什么这行不通吗?我的数据库结构是:
页面|标题
index.php | index
对不起,我不清楚。 $ page ['title'] =类型的resource(5)
的 dump 变量(mysql结果)。我怎样才能解决这个问题?谢谢您的帮助。
解决方案
mysql_query
返回一个资源,您需要使用 mysql_fetch_array
(或类似名称,例如 mysql_fetch_assoc
)以获取标题值。
例如:
$ res = mysql_query( SELECT title FROM页面WHERE page ='index.php');
$ page = mysql_fetch_assoc($ res);
// $ page ['title']现在包含该值(假设数据库中有一个index.php页面)。
警告:您的查询容易受到SQL注入的攻击。</ p>
<?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>
Does somebody know why this doesn't work? My DB structure is:
page |title
index.php|index
Sorry but I can't get this clear. The var dump of $page['title'] = resource(5)
of type (mysql result). How can I fix this? Thanks for the help.
解决方案
mysql_query
returns a resource, you need to use mysql_fetch_array
(or similar, e.g. mysql_fetch_assoc
) on the resource to get the title value.
For example:
$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).
Warning: your query is vulnerable to SQL injection.
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