[英]Very complicated SQL query
我正在使用 MySql 并应该创建一个非常复杂的报告来显示事件状态。
例如:
event_id | state | create_time
----------------------------------------------------
1 | PENDING | 2019-04-21 12:55:59.312
1 | COMPLETED | 2019-04-21 12:55:59.339
2 | PENDING | 2019-04-21 11:40:21.699
3 | PENDING | 2019-04-21 11:40:21.699
3 | FAILED | 2019-04-21 11:40:21.600
3 | COMPLETED | 2019-04-21 11:40:21.578
我需要根据州偏好选择行:
这应该使用 OFFSET 和 SIZE 进行分页。
每个事件将显示一次。
预期结果:
event_id | state | create_time
----------------------------------------------------
1 | COMPLETED | 2019-04-21 12:55:59.339
2 | PENDING | 2019-04-21 11:40:21.699
3 | FAILED | 2019-04-21 11:40:21.600
如何在单个 SQL 查询中执行此操作。
问候,伊多
我怀疑您想要使用一些条件逻辑进行聚合,如下所示:
select event_id,
(case when sum(state = 'FAILED') > 0 then 'FAILED'
when sum(state = 'COMPLETED') > 0 then 'COMPLETED'
when sum(state = 'PENDING') > 0 then 'PENDING'
end) as new_state
from t
group by event_id;
您可以使用coalesce()
进行聚合:
select event_id,
coalesce(max(case when state = 'FAILED' then 'FAILED' end),
max(case when state = 'COMPLETED' then 'COMPLETED' end),
'PENDING'
)
from table t
group by event_id;
另一种选择是将这些状态首选项作为位掩码处理。
注意此方法假定 event_id 没有重复的统计信息。
询问
SELECT
t.*
FROM (
SELECT
t.event_id
, SUM(bitmask_table.bitmask) AS total_bitmask
FROM (
SELECT
DISTINCT
state
, CASE
WHEN state = 'PENDING'
THEN 2
WHEN state = 'COMPLETED'
THEN 4
WHEN state = 'FAILED'
THEN 8
END AS bitmask
FROM
t
) AS bitmask_table
INNER JOIN
t
ON
t.state = bitmask_table.state
GROUP BY
t.event_id
) AS group_bitmasked
INNER JOIN
t
ON
group_bitmasked.event_id = t.event_id
AND
t.state = CASE
WHEN group_bitmasked.total_bitmask & 8 AND group_bitmasked.total_bitmask & 4 AND group_bitmasked.total_bitmask & 2
THEN 'FAILED'
WHEN group_bitmasked.total_bitmask & 4 AND group_bitmasked.total_bitmask & 2
THEN 'COMPLETED'
WHEN group_bitmasked.total_bitmask & 2
THEN 'PENDING'
END
结果
| event_id | state | create_time |
| -------- | --------- | ----------------------- |
| 1 | COMPLETED | 2019-04-21 12:55:59.339 |
| 2 | PENDING | 2019-04-21 11:40:21.699 |
| 3 | FAILED | 2019-04-21 11:40:21.600 |
看演示
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.