Mysql - 根据数学运算和 SUM() 函数对结果进行分组

发布于 2024-08-29 01:53:49 字数 945 浏览 2 评论 0原文

我有以下两个表...

表:room_type

 type_id    type_name   no_of_rooms     max_guests  rate    
  1          Type 1         15               2      1254    
  2          Type 2         10               1      3025

表:预订

reservation_id  start_date  end_date    room_type   booked_rooms    
       1        2010-04-12  2010-04-15      1            8 
       2        2010-04-12  2010-04-15      1            2 

现在... 我有这个查询

SELECT type_id, type_name
FROM room_type
WHERE id NOT IN (SELECT room_type
             FROM reservation
             WHERE start_date >= '$start_date'
             AND   end_date <= '$end_date')

该查询的作用是选择在开始日期和结束日期之间未预订的房间。

另外,正如您从预订表中看到的,我们还有“两个日期之间预订的房间数量”因素...

我需要将这个“两个日期之间预订的房间数量”因素也添加到查询...

查询应返回两个日期之间至少有一个房间空闲的房间类型。

我弄清楚了逻辑,但无法将其表示为查询......!你将如何做到这一点...?

感谢您的建议..!

I'm having the following two tables...

Table : room_type

 type_id    type_name   no_of_rooms     max_guests  rate    
  1          Type 1         15               2      1254    
  2          Type 2         10               1      3025

Table : reservation

reservation_id  start_date  end_date    room_type   booked_rooms    
       1        2010-04-12  2010-04-15      1            8 
       2        2010-04-12  2010-04-15      1            2 

Now... I have this query

SELECT type_id, type_name
FROM room_type
WHERE id NOT IN (SELECT room_type
             FROM reservation
             WHERE start_date >= '$start_date'
             AND   end_date <= '$end_date')

What the query does is it selects the rooms that are not booked between the start date and end date.

Also, as you can see from the reservation table, we also have 'number of rooms booked between the two dates' factor also...

I need to add this 'no.of booked rooms between the two dates' factor also in to the query...

The query should return the type of rooms for which at least one room is free between the two dates.

I worked out the logic but just can't represent it as a query....! How will you do this...?

Thanks for your suggestions..!

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烈酒灼喉 2024-09-05 01:53:49

你的结果可能会在这个

 SELECT a.type_id, a.type_name, a.no_of_rooms, (

SELECT SUM( booked_rooms )
FROM reservation
WHERE room_type = a.type_id
AND start_date >= '2010-04-12'
AND end_date <= '2010-04-15'
) AS booked_rooms, (
a.no_of_rooms - (
SELECT SUM( booked_rooms )
FROM reservation
WHERE room_type = a.type_id
AND start_date >= '2010-04-12'
AND end_date <= '2010-04-15' )
) AS freerooms
FROM room_type AS a
LEFT JOIN reservation AS b ON a.type_id = b.room_type
GROUP BY a.type_id
ORDER BY a.type_id

Your result may be got in this

 SELECT a.type_id, a.type_name, a.no_of_rooms, (

SELECT SUM( booked_rooms )
FROM reservation
WHERE room_type = a.type_id
AND start_date >= '2010-04-12'
AND end_date <= '2010-04-15'
) AS booked_rooms, (
a.no_of_rooms - (
SELECT SUM( booked_rooms )
FROM reservation
WHERE room_type = a.type_id
AND start_date >= '2010-04-12'
AND end_date <= '2010-04-15' )
) AS freerooms
FROM room_type AS a
LEFT JOIN reservation AS b ON a.type_id = b.room_type
GROUP BY a.type_id
ORDER BY a.type_id
~没有更多了~
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