case = 1 时的条件计数,DB2
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【中文标题】case = 1 时的条件计数,DB2【英文标题】:Conditional count when case = 1, DB2 【发布时间】:2019-10-07 10:56:03 【问题描述】:我目前正在尝试找出在 DB2 for Iseries 中将条件计数作为别名的最佳方法。以下值表示可以创建、完成和取消作业的作业状态,因此任何一项作业都可能附加多个状态代码。
但是,对于我的最终价值,我正在尝试计算仅具有已创建状态的工作,以便我可以显示仍有多少工作处于开放状态。基本上寻找创建案例的计数= 1的案例,但以下在'='处失败
SELECT
COUNT(CASE A1.JOB WHEN = 'CREATED' THEN 1 END) AS CREATED,
COUNT(CASE A1.JOB WHEN = 'CANCELLED' THEN 1 END) AS CANCELLED,
COUNT(CASE WHEN A1.JOB 'CREATED' = 1 then 1 END) AS OPEN
FROM SCHEMA.TABLE A1;
样本数据和结果:
Job ID | Status_code
-------------------------
123 'CREATED'
123 'COMPLETED'
521 'CREATED'
521 'CANCELLED'
645 'CREATED'
结果:
JOB | CREATED | CANCELLED | OPEN
-------------------------------------------
123 1 0 0
521 1 1 0
645 1 0 1
【问题讨论】:
样本数据和期望的结果会有所帮助。 @GordonLinoff 抱歉,现在才添加 【参考方案1】:假设唯一的“关闭”状态是'CANCELLED'
,您可以像这样使用not exists
:
select count(*)
from schema.table t
where t.status_code = 'CREATED' and
not exists (select 1
from schema.table t2
where t2.job = t.job and
t2.status_code in ('CANCELLED', 'COMPLETED', 'DELETED')
);
如果您想要多个计数,则这样的过滤不起作用。所以先按作业汇总:
select sum(is_created) as num_created,
sum(is_cancelled) as num_cancelled,
sum(is_created * (1 - is_cancelled) * (1 - is_completed) * (1 - is_deleted)) as open
from (select job,
max(case when status_code = 'CREATED' then 1 else 0 end) as is_created,
max(case when status_code = 'CANCELLED' then 1 else 0 end) as is_cancelled,
max(case when status_code = 'COMPLETED' then 1 else 0 end) as is_completed,
max(case when status_code = 'DELETED' then 1 else 0 end) as is_deleted
from t
group by job
) j
【讨论】:
抱歉,我试图更好地澄清我的问题,但从技术上讲,它们可以作为结束状态完成、取消或删除 但是对于这个特定的查询,我只是想显示创建、取消和打开【参考方案2】:以下返回你需要的结果:
WITH TAB (Job_ID, JOB) AS
(
VALUES
(123, 'CREATED')
, (123, 'COMPLETED')
, (521, 'CREATED')
, (521, 'CANCELLED')
, (645, 'CREATED')
)
SELECT
Job_ID
, COUNT(CASE A1.JOB WHEN 'CREATED' THEN 1 END) AS CREATED
, COUNT(CASE A1.JOB WHEN 'CANCELLED' THEN 1 END) AS CANCELLED
, CASE
WHEN NULLIF(COUNT(1), 0) = COUNT(CASE A1.JOB WHEN 'CREATED' then 1 END)
THEN 1
ELSE 0
END AS OPEN
FROM TAB A1
GROUP BY JOB_ID;
【讨论】:
【参考方案3】:使用条件聚合:
SELECT
JobID,
MAX(CASE Status_code WHEN 'CREATED' THEN 1 ELSE 0 END) AS CREATED,
MAX(CASE Status_code WHEN 'CANCELLED' THEN 1 ELSE 0 END) AS CANCELLED,
MIN(CASE WHEN Status_code <> 'CREATED' THEN 0 ELSE 1 END) AS OPEN
FROM tablename
GROUP BY JobID
请参阅demo。 结果:
> JobID | CREATED | CANCELLED | OPEN
> ----: | ------: | --------: | ---:
> 123 | 1 | 0 | 0
> 521 | 1 | 1 | 0
> 645 | 1 | 0 | 1
【讨论】:
【参考方案4】:假设有效的关闭状态是“COMPLETED”或“CANCELLED”,您可以尝试以下 SQL。
SELECT
A1.JobID,
sum(CASE WHEN A1.Status_code = 'CREATED' THEN 1 ELSE 0 END) AS CREATED,
sum(CASE WHEN A1.Status_code = 'CANCELLED' THEN 1 ELSE 0 END) AS CANCELLED,
(
SUM(CASE WHEN A1.Status_code = 'CREATED' THEN 1 ELSE 0 END)
- sum(CASE WHEN A1.Status_code = 'CANCELLED' THEN 1 ELSE 0 END)
- sum(CASE WHEN A1.Status_code = 'COMPLETED' THEN 1 ELSE 0 END)
) AS OPEN
FROM SCHEMA.TABLE A1
GROUP BY A1.JobID
【讨论】:
谢谢!这是给我正确数据的唯一选项,所以我接受了它 @TomN。 唯一选项是什么意思? @TomN。您接受了一个错误的答案,因为它会在为相同的jobid竞争、取消和创建时返回-1
:dbfiddle.uk/…
@forpas 好点,但它是给我适当数据的唯一解决方案,但不应该有这些情况。是否有解决 -1 问题的好方法?
这是给我适当数据的唯一解决方案你在说什么?我在答案中发布了一个小提琴演示,您可以在其中看到它返回正确的结果。以上是关于case = 1 时的条件计数,DB2的主要内容,如果未能解决你的问题,请参考以下文章