我有 2 个单独运行的查询,但不知道如何将它们合并为一个。
第一个计算一个植物的可用性,如下所示:
SELECT *,
(off_time - on_time)/(UNIX_TIMESTAMP('x') - UNIX_TIMESTAMP('y')) AS A
FROM(
SELECT equipment_id,
IF(time_on < 'x', UNIX_TIMESTAMP('x'), UNIX_TIMESTAMP(time_on)) AS on_time,
(CASE
WHEN (time_off IS NULL) THEN UNIX_TIMESTAMP(NOW())
WHEN (time_off > 'y') THEN UNIX_TIMESTAMP('y')
ELSE UNIX_TIMESTAMP(time_off)
END) AS off_time
FROM r) AS T
GROUP BY equipment_id
但我只想显示被识别为可用于服务的工厂,该工厂存储在单独的表中,因此我认为我必须使用 JOIN,例如:
SELECT r.equipment_id, r.time_on, r.time_off, i.in_service
FROM r
INNER JOIN i
ON r.equipment_id=i.equipment_id
但是,我尝试将两者结合起来却失败了。我是自学成才,对此非常陌生,因此任何帮助、评论甚至只是对我的疑问的一些批评都是值得赞赏的。
最佳答案
那么你可以像这样加入它。
SELECT
*,
((off_time - on_time)/(UNIX_TIMESTAMP('x') - UNIX_TIMESTAMP('y'))) AS A
FROM (SELECT
r.equipment_id,
i.in_service,
IF(time_on < 'x', UNIX_TIMESTAMP('x'), UNIX_TIMESTAMP(time_on)) AS on_time,
(CASE WHEN (time_off IS NULL) THEN UNIX_TIMESTAMP(NOW()) WHEN (time_off > 'y') THEN UNIX_TIMESTAMP('y') ELSE UNIX_TIMESTAMP(time_off) END) AS off_time
FROM r
left join i
ON r.equipment_id = i.equipment_id) AS T
GROUP BY equipment_id
关于MYSQL 内连接和派生表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15066615/