-2

这是我的代码

<?php

session_start();

include 'header.php';

$getUserInfo = mysql_query("SELECT * FROM Campaigns WHERE cid = '$cid'");
$userinfo = mysql_fetch_assoc($getUserInfo);

$cid = $getUserInfo['cid']; //ID
$CName = $getUserInfo['CName']; //Name
$CDesc = $getUserInfo['CDesc']; //Description
$CAmt = $getUserInfo['CAmt']; //Rate
$CReqs = $getUserInfo['CReqs']; //Requirements 
?>
<html>
    <body>

<h3><?php echo $cid;?>, <?php echo $CName;?></h3>
<p><?php echo $CDesc;?>     

    </body>
</html>

对于 php 数据库中的每个新项目，我希望它能够动态显示。

但由于某种原因，没有显示任何信息。是因为我正在使用 getUserInfo

4

1 回答 1

2

代替

$cid = $getUserInfo['cid']; //ID
$CName = $getUserInfo['CName']; //Name
$CDesc = $getUserInfo['CDesc']; //Description
$CAmt = $getUserInfo['CAmt']; //Rate
$CReqs = $getUserInfo['CReqs']; //Requireme

和

$cid = $userinfo ['cid']; //ID
$CName = $userinfo ['CName']; //Name
$CDesc = $userinfo ['CDesc']; //Description
$CAmt = $userinfo ['CAmt']; //Rate
$CReqs = $userinfo ['CReqs']; //Requireme

您正在从错误的变量中读取

进一步的建议：

确保您确实从 MySQL 获得了结果：

var_dump($getUserInfo);
var_dump($userinfo);

如果没有，请检查为什么不使用echo mysql_error();

确保数组中有正确的键（PHP 变量区分大小写）

于 2013-09-21T17:16:11.097 回答