2013-09-21 51 views
-2

這裏是我的代碼顯示動態信息PHP的

<?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> 

而且在PHP數據庫中的每個新項目我希望它可以動態顯示。

但由於某種原因沒有顯示信息。難道是因爲我使用getUserInfo

+2

它不應該是'$ CID = $用戶信息[ 'CID'];'? – TheNytangel

回答

2

更換

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

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

你是從錯誤的變量讀

進一步建議:

請務必切實從MySQL獲取結果:

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

如果沒有,檢查爲什麼不echo mysql_error();

確保你有你的數組中的右方向鍵(PHP變量是區分大小寫)

+1

另外,不要使用'mysql_'函數,特別是如果你正在學習。用'mysqli_''或PDO對象開始吧。 –

+0

當我用代碼替換它沒有工作,我得到了相同的結果沒有工作。並感謝我將取代mysqli_ – user2802562