当我运行我的代码
时,我总是会收到此错误0:
Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\fyp\admin_vieworders_details.php on line 12
Query Failed!You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
这是我的数据库
table order_detail {Order_Detail_ID(PK),Order_Quantity,Sub_Total}
表单{Order_ID,Order_Date,Order_Time,Delivery_Charge,Delivery_Name}
这是我的代码
$orderid =@$_GET["id"];
$detail = "SELECT
order.*,
order_detail.Order_Quantity,
order_detail.Sub_Total
FROM `order` AS a, `order_detail` AS b
WHERE order_detail.Order_ID = `order`.Order_ID AND order_detail.Order_Detail_ID=$orderid";
$result = mysql_query($detail);
$row = mysql_fetch_assoc($result);
if($result === FALSE)
{
die("Query Failed!".mysql_error().$result);
}
答案 0 :(得分:1)
在致电mysql_error()
之前检查mysql_fetch_assoc()
。您的查询失败,查询函数返回false
。
$result = mysql_query($detail);
if (!$result) {
die("Error: ".mysql_error()); // Note: raw database errors are useless for users
// better log the error an create a "nice" error page
}
$row = mysql_fetch_assoc($result);
或者这样可能会让你开始。
答案 1 :(得分:1)
您的查询中似乎有错误:
尝试:
$detail = "SELECT
a.*,
b.Order_Quantity,
b.Sub_Total
FROM `order` AS a, `order_detail` AS b
WHERE b.Order_Detail_ID = a.Order_ID AND
b.Order_Detail_ID=$orderid";
而不是
$detail = "SELECT
order.*,
order_detail.Order_Quantity,
order_detail.Sub_Total
FROM `order` AS a, `order_detail` AS b
WHERE order_detail.Order_ID = `order`.Order_ID AND
order_detail.Order_Detail_ID=$orderid";
还有......
$orderid =$_GET["orderid"];
而不是
$orderid =@$_GET["id"];
答案 2 :(得分:0)
更改此
$orderid =@$_GET["id"];
到
$orderid =$_GET['id'];