SQL: 最旧项目的年龄

Question

我有以下 Oracle 数据库。某个 ID 可以具有已接受状态，但可以在失败状态下继续，然后 return 回到已接受状态。这可能发生 X 次，所以它可以被称为循环

ID       STATUS         TIME INSERT
------------------------------
1        Accepted       01:00:00
1        Failed         02:00:00
1        Accepted       02:30:00
2        Accepted       02:33:00

我想计算具有已接受状态的最旧项目的年龄。这意味着 SYSDATE - TIME INSERT

的结果

As Result of this data I expect the 3d record time value, 02:30:00

我需要检查每个 ID 的最年轻时间，以及 ID 的所有最年轻生命周期中最老的生命周期。这对我来说看起来很复杂。

工作流程步骤为：

状态为Accepted的特定id的最小时间值
第 1 步针对每个处于已接受状态的 ID
我在步骤 2 中计算的所有最小值的最大值。

此时我有以下代码

select min(TO_CHAR(TO_DATE('20000101','yyyymmdd')+(SYSDATE - dm1.time_insert),'hh24:mi:ss'))
from db dm1 
where dm1.status='accepted'
group by dm1.id;

这为我提供了所有记录的最小值序列。但是现在我需要所有这些最小值中的最大值，我该怎么做？谁能破案？

Answer 1

希望对您有所帮助：

-- this will give you the minimum date of all id's
select * from tab1 where TIME_INSERT  in (select min(TIME_INSERT) from tab1 )and STATUS ='Accepted'

-- this will give you the minimum date of a specefique id 
select * from tab1 where TIME_INSERT  in (select min(TIME_INSERT) from tab1 )
and STATUS ='Accepted' and ID =1

检查这个

select id, TIME_INSERT from
 (select TIME_INSERT, ID , min(TIME_INSERT) over (partition by ID) maxid from tab1) 
where TIME_INSERT = maxid and TIME_INSERT in(select max(TIME_INSERT) from tab1) group by ID,TIME_INSERT

Answer 2

您可以只查找 max(time_insert) 并按 id 分组。查看示例查询

    with src as (
select 1 as id, 'Accepted' as status, to_date('2015-05-05 01:00:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual
union all
select 1 as id, 'Failed' as status, to_date('2015-05-05 02:00:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual
union all 
select 1 as id, 'Accepted' as status, to_date('2015-05-05 02:30:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual
union all
select 2 as id, 'Accepted' as status, to_date('2015-05-05 02:33:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual)

select id, max(time_insert) as time_insert, max(sysdate - time_insert) as lifetime_max, min(sysdate - time_insert) as liftime_min from src 
where status ='Accepted'
group by id

Edit: Is this what you want to achieve:

with src as (
select 1 as id, 'Accepted' as status, to_date('2015-05-05 01:00:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual
union all
select 1 as id, 'Failed' as status, to_date('2015-05-05 02:00:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual
union all 
select 1 as id, 'Accepted' as status, to_date('2015-05-05 02:30:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual
union all
select 2 as id, 'Accepted' as status, to_date('2015-05-05 02:33:00','yyyy-mm-dd hh24:mi:ss') as time_insert from dual)
, src2 as (
select max(time_insert) as time_insert, max(sysdate - time_insert) as lifetime_max, min(sysdate - time_insert) as lifetime_min from src 
where status ='Accepted')
select max(lifetime_min) from src2

Answer 3

这样就可以了。

SELECT id,
       min(time_insert),
       max(time_insert)
FROM   
       your_table t1
WHERE  
       status = 'Accepted' 
AND    time_insert > (
          SELECT
                  nvl(max(time_insert), to_date('01-JAN-1970', 'DD-MON-YYYY'))
          FROM 
                  your_table
          WHERE 
                  status = 'Failed'
          AND     id = t1.id
       )
GROUP BY id

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