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Php 显示动态信息_Php_Database - Fatal编程技术网
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Php 显示动态信息

php database

Php 显示动态信息,php,database,Php,Database,这是我的密码 <?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

这是我的密码

<?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>

替换


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


与

您正在读取错误的变量

进一步建议：

确保您确实从MySQL获得了一个结果：


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


如果没有，请使用
echo mysql_error（）检查为什么没有


确保数组中有正确的键（PHP变量区分大小写）
是否应该是
$cid=$userinfo['cid']？另外，不要使用
mysql\uuz
函数，尤其是在学习时。从
mysqli\uu
或PDO对象开始。当我用代码替换它时，它不起作用，我得到了相同的结果，什么都不起作用。谢谢，我将用mysqli替换_

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