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Php 资源id#5:MySQL_Php_Mysql_Database - Fatal编程技术网

Php 资源id#5:MySQL

Php 资源id#5:MySQL,php,mysql,database,Php,Mysql,Database,返回一个资源,您需要在该资源上使用(或类似的,例如)来获取标题值 例如: <?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 = mys

返回一个资源,您需要在该资源上使用(或类似的,例如)来获取标题值

例如:

<?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>
警告:您的查询易受SQL注入攻击。

返回资源,您需要在资源上使用(或类似的,例如)来获取标题值

例如:

<?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>

警告:您的查询易受SQL注入攻击。

这是由
mysql\u query
对资源使用
mysql\u fetch\u assoc
返回的资源id,以获取标题值

执行以下更改:

$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).

这是由
mysql\u query
对资源使用
mysql\u fetch\u assoc
返回的资源id,以获取标题值

执行以下更改:

$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).

您需要使用类似于
mysql\u fetch\u assoc()
mysql\u result()
的函数来获取实际数据

 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'];

您需要使用类似于
mysql\u fetch\u assoc()
mysql\u result()
的函数来获取实际数据

 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'];

这项服务迟了4年才推出,但mysql已被弃用。您应该改用mysqli_*或PDO函数

这比4年前晚了,但mysql_*已被弃用。您应该改用mysqli_*或PDO函数

mysql_fetch_数组
鼓励潜在的不良做法。使用
mysql\u fetch\u assoc
更好(也更有效),不过在这种情况下,您只需要从一行中获取一个字段,我会使用
mysql\u result
。现在就可以了!谢谢大家的快速回答。(我使用了mysql\u result($result,这里是0.=)
mysql\u fetch\u数组
鼓励潜在的不良做法。使用
mysql\u fetch\u assoc
更好(也更有效),不过在这种情况下,您只需要从一行中获取一个字段,我会使用
mysql\u result
。现在就可以了!谢谢大家的快速回答。(我使用了mysql_result($result,这里为0.=))