Question

我有一个包含事件的数据库表：

CREATE TABLE events
(event VARCHAR(32)
,down_time TIMESTAMP
,up_time TIMESTAMP
,id INT UNSIGNED NOT NULL AUTO_INCREMENT KEY
,INDEX(event(16))
);

1   2   3   4   5   6   7   8   9   10
|   |   |   |-------e1------|   |   |
|   |   |---e2--|   |-------e2------|   
|   |-------------e3------------|   |
|----------e4-----------|   |   |   |

INSERT INTO events(event, down_time, up_time) VALUES
    ('e1', '2015-01-01 00:00:04', '2015-01-01 00:00:08'),
    ('e2', '2015-01-01 00:00:03', '2015-01-01 00:00:05'),
    ('e2', '2015-01-01 00:00:06', '2015-01-01 00:00:10'),
    ('e3', '2015-01-01 00:00:02', '2015-01-01 00:00:09'),
    ('e4', '2015-01-01 00:00:01', '2015-01-01 00:00:07');

我希望找到它们相交的时间，并且能够通过组合两个查询来实现这一点。

第一次查询：

CREATE VIEW overlap1 AS
SELECT
CONCAT(a.event,'-', b.event) overlaps,
GREATEST(a.down_time,b.down_time) AS downtime, 
LEAST(a.up_time,b.up_time) AS uptime,
TIME_TO_SEC(TIMEDIFF( LEAST(a.up_time,b.up_time),
    GREATEST(a.down_time,b.down_time))) AS duration

FROM events a 
JOIN events b 
ON  a.event < b.event 
    AND (a.event = 'e1' OR a.event = 'e2' OR a.event = 'e3' OR a.event = 'e4')
    AND (b.event = 'e1' OR b.event = 'e2' OR b.event = 'e3' OR b.event = 'e4') 
WHERE   ( a.down_time <= b.up_time )
AND     ( a.up_time >= b.down_time );

SELECT * FROM overlap1;

+----------+---------------------+---------------------+----------+
| overlaps | downtime            | uptime              | duration |
+----------+---------------------+---------------------+----------+
| e1-e2    | 2015-01-01 00:00:04 | 2015-01-01 00:00:05 |        1 |
| e1-e2    | 2015-01-01 00:00:06 | 2015-01-01 00:00:08 |        2 |
| e1-e3    | 2015-01-01 00:00:04 | 2015-01-01 00:00:08 |        4 |
| e2-e3    | 2015-01-01 00:00:03 | 2015-01-01 00:00:05 |        2 |
| e2-e3    | 2015-01-01 00:00:06 | 2015-01-01 00:00:09 |        3 |
| e1-e4    | 2015-01-01 00:00:04 | 2015-01-01 00:00:07 |        3 |
| e2-e4    | 2015-01-01 00:00:03 | 2015-01-01 00:00:05 |        2 |
| e2-e4    | 2015-01-01 00:00:06 | 2015-01-01 00:00:07 |        1 |
| e3-e4    | 2015-01-01 00:00:02 | 2015-01-01 00:00:07 |        5 |
+----------+---------------------+---------------------+----------+

第二次查询：

CREATE VIEW overlap2 AS
SELECT
CONCAT(a.overlaps,'-',b.overlaps) AS overlaps,
GREATEST(a.downtime,b.downtime) AS downtime,
LEAST(a.uptime,b.uptime) AS uptime,
TIMEDIFF( LEAST(a.uptime,b.uptime),
GREATEST(a.downtime,b.downtime) ) AS duration

FROM overlap1 a
JOIN overlap1 b 
ON a.overlaps < b.overlaps
AND (a.overlaps = 'e1-e2' OR a.overlaps = 'e3-e4')
AND (b.overlaps = 'e1-e2' OR b.overlaps = 'e3-e4')

WHERE   ( a.downtime <= b.uptime )
AND ( a.uptime >= b.downtime );

SELECT * FROM overlap2;

+-------------+---------------------+---------------------+----------+
| overlaps    | downtime            | uptime              | duration |
+-------------+---------------------+---------------------+----------+
| e1-e2-e3-e4 | 2015-01-01 00:00:04 | 2015-01-01 00:00:05 | 00:00:01 |
| e1-e2-e3-e4 | 2015-01-01 00:00:06 | 2015-01-01 00:00:07 | 00:00:01 |
+-------------+---------------------+---------------------+----------+

如果可能的话，我想使用一个查询来实现这一点。非常感谢！

mysql将两个查询合并为一个

0 个答案: