Question

我有以下mysql表（房间）

id_user_room     username   room_name   date_from        date_to 
    1             mike        A1        2017-02-01       2017-02-10
    2             evans       A2        2017-01-20       2017-01-30
    3             mike        A3        2017-02-16       2017-02-20
    4             mike        A1        2017-03-01       2017-03-18

如果我想知道2017-01-01和2017-05-30之间迈克的房间，我执行以下SQL查询。

SELECT room_name, date_from, date_to
FROM rooms
WHERE (date_to >= '2017-01-01' AND date_from <= '2017-05-30')
  AND username='Mike'
order by date_to ASC

此查询提供以下结果：

room_name     date_from      date_to
A1           2017-02-01     2017-02-10
A3           2017-02-16     2017-02-20
A1           2017-03-01     2017-03-18

有没有办法获得其余（相反）的日期范围？我的意思是我想知道迈克没有住在房间里的范围。对于具体示例，我想得到的结果将是以下日期范围：

2017-01-01 to 2017-01-31
2017-02-11 to 2017-02-15
2017-02-21 to 2017-02-28
2017-03-19 to 2017-05-30

非常感谢你的帮助!!!

Answer 1

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(username VARCHAR(12) NOT NULL
,room_name CHAR(2) NOT NULL
,date_from DATE
,date_to DATE
,PRIMARY KEY  (username,room_name,date_from)
);

INSERT INTO my_table VALUES
('mike','A1','2017-02-01','2017-02-10'),
('evans','A2','2017-01-20','2017-01-30'),
('mike','A3','2017-02-16','2017-02-20'),
('mike','A1','2017-03-01','2017-03-18');

     SELECT username
          , '2017-01-01' date_from
          , MIN(date_from - INTERVAL 1 DAY) date_to 
       FROM my_table 
      GROUP 
         BY username
      UNION
     SELECT x.username
         , x.date_to + INTERVAL 1 DAY
         , MIN(y.date_from - INTERVAL 1 DAY)
       FROM my_table x
       JOIN my_table y ON y.date_to > x.date_to
        AND y.username = x.username
      GROUP 
         BY x.username
          , x.date_to
      UNION
     SELECT username
          , MAX(date_to + INTERVAL 1 DAY) 
          , '2017-05-30'
       FROM my_table 
      GROUP 
         BY username;
+----------+------------+------------+
| username | date_from  | date_to    |
+----------+------------+------------+
| evans    | 2017-01-01 | 2017-01-19 |
| mike     | 2017-01-01 | 2017-01-31 |
| mike     | 2017-02-11 | 2017-02-15 |
| mike     | 2017-02-21 | 2017-02-28 |
| evans    | 2017-01-31 | 2017-05-30 |
| mike     | 2017-03-19 | 2017-05-30 |
+----------+------------+------------+
6 rows in set (0.01 sec)

这假设所有日期都在阈值范围内 - 但如果他们不这样做，则不是特别复杂的调整。

Answer 2

创建房间表，如上所述：

CREATE TABLE rooms (
    id_user_room INTEGER,
    username     CHAR(10),
    room_name    CHAR(10),
    date_from    TIMESTAMP,
    date_to      TIMESTAMP
);

INSERT INTO rooms (id_user_room, username, room_name, date_from, date_to) VALUES
(1, 'mike', 'A1', '2017-02-01', '2017-02-10'),
(2, 'evans', 'A2', '2017-01-20', '2017-01-30'),
(3, 'mike', 'A3', '2017-02-16', '2017-02-20'),
(4, 'mike', 'A1', '2017-03-01', '2017-03-18')
;

你必须创建一个类似的辅助日期表：

CREATE TABLE dates (
    day      TIMESTAMP
);

INSERT INTO dates (day) VALUES
('2017-01-01'),
('2017-01-02'),
('2017-01-03'),
...
('2017-12-29'),
('2017-12-30'),
('2017-12-31')
;

有了这个，您可以找到迈克没有住在房间里的日期：

CREATE TABLE empty_dates AS
SELECT
    d.day
FROM
    dates d
LEFT JOIN
    rooms r ON r.username = 'mike' AND d.day BETWEEN r.date_from AND r.date_to
WHERE
  r.id_user_room IS NULL
;

为了简洁起见，我实现了这个查询而不是使用子查询。在其他SQL语言中，比如TSQL或PostgreSQL，我宁愿使用CTE。现在您可以指定间隔＆＃39;将此表连接到自身的开始日期和结束日期。

SELECT 
    CASE WHEN ed3.day IS NULL THEN 'Start date' ELSE 'End date' END row_type,
    ed1.day
FROM
    empty_dates ed1
LEFT JOIN
    empty_dates ed2 ON DATEDIFF(ed2.day, ed1.day) = 1
LEFT JOIN
    empty_dates ed3 ON DATEDIFF(ed3.day, ed1.day) = -1
WHERE
    -- the next date is in the rooms table
    ed2.day IS NULL
    -- the previous date is in the rooms table
    OR ed3.day IS NULL
;

如何在重叠的日期范围之间找到其余部分

2 个答案: