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我被困在試圖使下面的一段代碼看起來似乎我的語法中的東西沒有正確聲明我試圖改變變量的名稱,但仍然沒有工作,我在我的數據庫中有一個名爲包的表,有什麼錯誤可能是什麼想法?請:(:PHP/MYSQL使用數據庫和PHP
<?PHP
include('connection.php');
include('validation.php');
include('header.php');
$PID = $_GET['PID'];
$q_getpackage = "SELECT * FROM packages WHERE PID=$PID";
$r_getpackage = mysql_query($q_getpackage,$connection);
?>
<!DOCTYPE HTML>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<link href="styles.css" type="text/css" rel="stylesheet"/>
<link href='http://fonts.googleapis.com/css?family=Courgette' rel='stylesheet' type='text/css'>
<link href='http://fonts.googleapis.com/css?family=Numans' rel='stylesheet' type='text/css'>
<title>Products Information</title>
</head>
<body>
<div class="packages_wraper"><!--Packages Wraper-->
<div class="package_name"><?php echo mysql_result($r_getpackage,0,'name');?></h3>
<div class="details"><?php echo mysql_result($r_getpackage,0,'details');?</div>
<div><?php echo mysql_result($r_getpackage,0,'guide');?> </div>
<div><?php echo mysql_result($r_getpackage,0,'review');?> </div>
<div class="pic"><img src="<?php echo mysql_result($r_getpackage,0,'pic_url');?>"/></div>
</div><!--END Packages Wraper-->
</div>
</body>
</html>
我有一個錯誤,指出:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/tk1/domains/valbet.tk/public_html/product_info.php on line 33
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/tk1/domains/valbet.tk/public_html/product_info.php on line 34
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/tk1/domains/valbet.tk/public_html/product_info.php on line 35
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/tk1/domains/valbet.tk/public_html/product_info.php on line 36
嘗試'0 1 2 3',而不是'「名」,「細節」,「G使用者「和」評論「。順便說一句,這將只顯示查詢的第一個結果 –