1

我在一张桌子上有一份办公室清单，以及每个办公室拥有的所有短期租约的清单。

我正在尝试查询这两个表以显示所有办公室的列表，以及预订的最早租约和再次可用时的最后租约。

SELECT offices.* ,MIN(lease.date_start), MAX(lease.date_end) FROM offices, lease WHERE lease.office_id = office.id ORDER BY office.id DESC

办公桌：

id        | office_name | office_description 
1         | North York  | Lorem Ipsum
2         | Toronto     | Lorem Ipsum
3         | Richmond    | Lorem Ipsum

租赁表：

id        | office_id   | start_date   | end_date 
1         | 1           | 5            | 8
2         | 1           | 3            | 7 
3         | 2           | 1            | 4

我试图得到的结果：

office_id=>1, start_date=>3, end_date=>8
office_id=>2, start_date=>1, end_date=>4
office_id=>3, start_date=>NULL, end_date=>NULL

我将如何构建我的查询以获得该结果？

4

4 回答 4

2

为了正确使用MIN()，MAX()在这种情况下，您需要GROUP BY：

SELECT
    office.*
    MIN(lease.start_date),
    MAX(lease.end_date)
FROM office AS office
LEFT JOIN lease AS lease ON (lease.office_id = office.office_id)
GROUP BY office.office_id
ORDER BY office.office_id DESC
于 2013-10-23T20:20:26.717 回答
0 
SELECT CONCAT("office_id=>",O.id,", start_dt=>",B.start_dt,", end_dt=>",B.end_dt)
FROM
office O,
(
 SELECT L.office_id,MIN(start_dt) AS start_dt,MAX(end_dt) AS end_dt
 FROM Lease L
 GROUP BY L.office_id
) B
WHERE O.id = B.office_id;
于 2013-10-23T20:21:07.050 回答
0

以下选择应为您提供所需的结果：

SELECT o.*, min(l.date_start), max(l.date_end)  
FROM offices o
LEFT JOIN lease l on (
   l.office_id = o.id
)
GROUP BY o.id
于 2013-10-23T20:26:58.937 回答
0 
SELECT office_id,
MIN(start_date) as start_date,
MAX(end_date) as end_date
FROM office 
LEFT JOIN lease ON (lease.office_id = office.office_id)
GROUP BY office_id
ORDER BY office_id ASC

或者

SELECT office_id, start_date, end_date FROM (
SELECT office_id,
MIN(start_date) as start_date,
MAX(end_date) as end_date
FROM lease GROUP BY office_id
UNION
SELECT id AS office_id,
NULL start_date, NULL end_date
FROM office 
WHERE NOT EXISTS 
(SELECT 1 FROM lease
WHERE office_id=office.id) ) u
ORDER BY office_id ASC

我不确定哪个会更快。

于 2013-10-23T20:31:30.467 回答