Php 警告:mysql_fetch_assoc()希望参数1是资源,布尔值在中给出

Php 警告:mysql_fetch_assoc()希望参数1是资源,布尔值在中给出,php,mysql,sql,Php,Mysql,Sql,当我运行我的代码时,我总是遇到这个错误 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

当我运行我的代码时,我总是遇到这个错误

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

表订单详细信息{订单详细信息ID(PK),订单数量,小计}

表订单{订单ID、订单日期、订单时间、交货费用、交货名称}

这是我的密码

$orderid =@$_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);
    }
调用
mysql\u-fetch\u-assoc()
之前,请检查
mysql\u-error()
。您的查询失败,查询函数返回了
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);

或者这样可能会让您开始。

您的查询似乎有错误:

尝试:

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";
$orderid =@$_GET["id"]; 
而且

$orderid =$_GET["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";
$orderid =@$_GET["id"]; 
改变这个

 $orderid =@$_GET["id"]; 


警告:
mysql.*
函数从PHP 5.5.0开始就不推荐使用,将来将被删除。相反,应该使用or扩展名。是否在此查询之前尝试过echo
$orderid
,如果它包含值?@Jess Tan:说真的,请查看标记为“相关”的右侧列。我们知道每个人的问题都是特殊的,但这只是一个常见的错误。你的问题中有一个错误。我经常看到这种类型的问题,你应该阅读。