编写满足两种情况的单个 SQL 查询
[英]Writing single SQL query satisfying two cases
问题描述
收到评论后，我将用所需的数据重新表述这个问题。
Vertica 分析函数： https : //www.vertica.com/blog/analytic-queries-vertica/
表格1：
create table etl_group_membership
(
group_item_id int not null,
member_item_id int not null
);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335640, 117722);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335640, 104151);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335640, 5316);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335641, 117723);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335641, 104152);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335641, 5317);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335642, 117724);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335642, 104153);
INSERT INTO etl_group_membership (group_item_id, member_item_id) VALUES (335642, 5318);
表 2：
create table v_poll_item
(
device_item_id int not null,
item_id int not null
);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (117722, 273215);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (117722, 117936);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (117722, 117873);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (117722, 123305);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (104151, 240006);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (104151, 240005);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (104151, 239415);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (104151, 239414);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (5316, 118310);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (5316, 130627);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (5316, 298564);
INSERT INTO v_poll_item (device_item_id, item_id) VALUES (5316, 118311);
表 3：注意im_utilization 也可以为 NULL
create table nrm_cpustats_rate
(
item_id int not null,
tstamp datetime not null,
im_utilization float,
);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (273215, '2021-06-28 19:55:00.000000', 2);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (273215, '2021-06-27 23:35:00.000000', 24);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (273215, '2021-06-26 14:05:00.000000', 27);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (273215, '2021-06-25 09:05:00.000000', 29);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-28 19:30:00.000000', 17);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-28 19:15:00.000000', 35);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-28 19:05:00.000000', 50);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-27 05:45:00.000000', 89);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-25 09:20:00.000000', 37);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-25 09:10:00.000000', 51);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (117936, '2021-06-25 08:50:00.000000', 90);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (118310, '2021-06-23 04:10:00.000000', 51);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (118310, '2021-06-23 03:15:00.000000', 48);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (118310, '2021-06-22 22:20:00.000000', 19);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (239414, '2021-06-22 17:10:00.000000', 11);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (239414, '2021-06-22 16:30:00.000000', 37);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (239414, '2021-06-22 16:35:00.000000', 38);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (239414, '2021-06-28 18:45:00.000000', 74);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (239414, '2021-06-28 18:48:00.000000', 76);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (239414, '2021-06-28 18:50:00.000000', 77);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (118311, '2021-06-28 00:40:00.000000', 29);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (118311, '2021-06-23 22:30:00.000000', 37);
INSERT INTO nrm_cpustats_rate (item_id, tstamp, im_utilization) VALUES (118311, '2021-06-23 22:25:00.000000', 92);
要获取组中的设备项 ID：
SELECT member_item_id FROM etl_group_membership WHERE group_item_id = 335640;
从检索到的设备项 id 列表中，获取 item_id 列表：
SELECT item_id FROM v_poll_item WHERE device_item_id IN (<devices retrieved from previous query>);
输入：
- 两个时间范围：昨天到 7 天前（
AND tstamp > '2021-06-22 00:00:00.000000'AND tstamp <= '2021-06-28 23:59:59.000000'） - 群号：335640
- 违约阈值：25
- 每天最少违规次数：2
预期输出：
违反意味着 im_utilization >= 25
仅选择在给定日期 count_of_breached >= 2 的那些记录，即排除 item_id 273215 的记录，因为即使违规次数 (>= 25) 为 2，每天也只有一个
device_item_id | item_id | count_of_breach | date_when_breached | max_utilization | max_utilization_tstamp
=====================================================================================================================
117722 | 117936 | 2 | 2021-06-28 | 90 | 2021-06-25 08:50:00.000000
117722 | 117936 | 3 | 2021-06-25 | 90 | 2021-06-25 08:50:00.000000
5316 | 118310 | 2 | 2021-06-23 | 51 | 2021-06-23 04:10:00.000000
5316 | 118311 | 2 | 2021-06-23 | 92 | 2021-06-23 22:25:00.000000
104151 | 239414 | 2 | 2021-06-22 | 77 | 2021-06-28 18:50:00.000000
104151 | 239414 | 3 | 2021-06-28 | 77 | 2021-06-28 18:50:00.000000
即使无法编写单个 SQL 查询来生成此输出，也可以建议两个优化的查询。 @marcothesane 指出，即使是获取每日违规的查询也可以用更好的方式编写。
更新：
这个查询对我有用，可以在最大时找到最大 im_utlization 和 tstamp。 我不确定为什么我必须在两个地方使用时间戳范围！
SELECT t.item_id, t.tstamp, t.im_utilization
FROM (
SELECT item_id, MAX(im_utilization) AS max_cpu
FROM nrm_cpustats_rate
WHERE item_id IN (SELECT item_id from v_poll_item WHERE device_item_id IN (SELECT item_id FROM v_poll_item WHERE device_item_id IN (117722, 104151, 5316)))
AND tstamp > '2021-06-22 00:00:00.000000
AND tstamp <= '2021-06-28 23:59:59.000000'
GROUP BY item_id
) AS m
INNER JOIN nrm_cpustats_rate AS t
ON t.item_id = m.item_id
AND t.im_utilization = m.max_cpu
AND tstamp > '2021-06-22 00:00:00.000000
AND tstamp <= '2021-06-28 23:59:59.000000'
ORDER BY 1, 2 DESC, 3 DESC
2 个解决方案
解决方案1
0 2021-06-29 17:40:55
首先，您所有的初步工作都专门依赖于 cpstats 费率表。 但正如您所提到的，违规考虑是基于同一天超过 1 次（因此为 2 次或更多）。 因此，对于每个项目，每天（因此日期（ ncr.tstamp ）字段只获取 DATE 部分而不管时间如何，我们可以应用 HAVING 子句，以便只有按项目和日期分组的那些有多个违规行为。
然后，我们可以从中加入特定设备 ID 的轮询项表。 以下应该适合您。
select
vpi.device_item_id,
ncr.item_id,
count(*) count_of_breach,
date( ncr.tstamp ) date_when_breached,
max( ncr.im_utilization ) max_utilization
from
nrm_cpustats_rate ncr
join v_poll_item vpi
on ncr.item_id = vpi.item_id
where
ncr.im_utilization >= 25
group by
vpi.device_item_id,
ncr.item_id,
date( ncr.tstamp )
having
count(*) > 1
反馈 - 修改
从查询中删除第二个表并尝试在查询中使用 TRUNC() 函数而不是 date()。 还添加了您的日期过滤器限制。
select
ncr.item_id,
count(*) count_of_breach,
trunc( ncr.tstamp, 'DD' ) date_when_breached,
max( ncr.im_utilization ) max_utilization
from
nrm_cpustats_rate ncr
where
ncr.tstamp > '2021-06-22 00:00:00.000000'
AND ncr.tstamp <= '2021-06-28 23:59:59.000000'
AND ncr.im_utilization >= 25
group by
ncr.item_id,
trunc( ncr.tstamp, 'DD' )
having
count(*) > 1
解决方案2
0 2021-06-30 17:21:05
您可以使用::DATE操作将 TIMESTAMP 硬转换为 DATE。
正如其他人所做的那样：
您可以获取分组查询中的所有前五列，按device_item_id,v_poll_item.item_id,tstamp::DATE分组。 我把它作为一个通用的表表达式 (CTE) 放入 WITH 子句中。 然后，我用nrm_cpustats_rate对item_id和对先前获得的max_utilization和im_utilization的等谓词加入了该 CTE； 我最终过滤掉少于 2 个违规计数的行，并按时间戳范围再次过滤。
WITH
grp AS (
SELECT
device_item_id
, v_poll_item.item_id
, SUM(
CASE
WHEN im_utilization >= 25 THEN 1
ELSE 0
END
) AS count_of_breach
, tstamp::DATE as date_when_breached
, MAX(im_utilization) AS max_utilization
FROM nrm_cpustats_rate
JOIN v_poll_item USING(item_id)
JOIN etl_group_membership ON member_item_id=device_item_id
WHERE tstamp > '2021-06-22 00:00:00.000000'AND tstamp <= '2021-06-28 23:59:59.000000'
GROUP BY
device_item_id
, v_poll_item.item_id
, tstamp::DATE
)
SELECT
grp.*
, o.tstamp AS max_utilization_tstamp
FROM grp
JOIN nrm_cpustats_rate o
ON o.item_id=grp.item_id
AND max_utilization=im_utilization
WHERE tstamp > '2021-06-22 00:00:00.000000'AND tstamp <= '2021-06-28 23:59:59.000000'
AND count_of_breach >= 2
;
-- out device_item_id | item_id | count_of_breach | date_when_breached | max_utilization | max_utilization_tstamp
-- out ----------------+---------+-----------------+--------------------+-----------------+------------------------
-- out 5316 | 118310 | 2 | 2021-06-23 | 51 | 2021-06-23 04:10:00
-- out 104151 | 239414 | 3 | 2021-06-28 | 77 | 2021-06-28 18:50:00
-- out 104151 | 239414 | 2 | 2021-06-22 | 38 | 2021-06-22 16:35:00
-- out 117722 | 117936 | 3 | 2021-06-25 | 90 | 2021-06-25 08:50:00
-- out 5316 | 118311 | 2 | 2021-06-23 | 92 | 2021-06-23 22:25:00
-- out 117722 | 117936 | 2 | 2021-06-28 | 50 | 2021-06-28 19:05:00
用于查找满足条件 C 且不满足 NOT C 的键的引用的 SQL 查询
[英]SQL query to find references of keys satisfying condition C and not satisfying NOT C
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