Question

大家好我是新手，我需要有关预订酒店的帮助，我有这张桌子：

id, idhotel, room, data_start, data_end, price

现在我需要计算同一天在多天之间搜索房间时的总价我正在尝试此查询，但结果为0。

$query = "SELECT room, SUM(price) FROM price WHERE idhotel='".$_GET['id']."' BETWEEN 'data_start' AND 'data_end'  
ORDER BY room";

有人可以帮助我吗？

Answer 1

我认为你需要这样的东西：

SELECT idhotel, room, price, (DATEDIFF(data_end, data_start) * price) as total FROM rooms

为了测试创建下一个表并填写它，请选择*：

create table rooms (id int not null primary key auto_increment, idhotel int, room int, data_start date, data_end date, price int);

+----+---------+------+------------+------------+-------+
| id | idhotel | room | data_start | data_end   | price |
+----+---------+------+------------+------------+-------+
|  1 |       1 |    1 | 2013-05-09 | 2013-05-12 |   100 |
|  2 |       1 |    1 | 2013-05-20 | 2013-05-20 |   100 |
+----+---------+------+------------+------------+-------+

使用下一个查询获取总费用（修改以防止丢失一天）：

SELECT idhotel, room, price, ((DATEDIFF(data_end, data_start) + 1) * price) as total FROM rooms;

结果：

+---------+------+-------+-------+
| idhotel | room | price | total |
+---------+------+-------+-------+
|       1 |    1 |   100 |   400 |
|       1 |    1 |   100 |   100 |
+---------+------+-------+-------+

扩展

如果您想获得期间的房间总和，请尝试使用以下代码：

SELECT 
    idhotel, room, SUM(total) as total_sum 
FROM 
(
    SELECT 
        idhotel, 
        room,
        price, 
        (DATEDIFF(data_end, data_start) * price) as total 
    FROM 
        rooms
) as t 
GROUP BY 
    idhotel, room;

结果：

+---------+------+-----------+
| idhotel | room | total_sum |
+---------+------+-----------+
|       1 |    1 |       150 |
+---------+------+-----------+

Answer 2

如果我理解正确你可能正在寻找类似的东西

$date_start = $_GET['date_start'];
$date_end = $_GET['date_end'];
$hotel_id = $_GET['id'];

$query = "SELECT room, 
                 SUM(price) total
            FROM tablename
           WHERE data_start >= '$date_start' 
             AND date_end <= '$date_end'
             AND idhotel = $hotel_id
           GROUP BY room";

计算不同日期之间的价格

2 个答案:

扩展