我需要编写Oracle sql查询以获得如下输出。开始和结束日期都是DATE数据类型。我有一个这样的文件
ApName DbName Status StartDate EndDate
A1 A11 Success 8/3/2016 8/3/2016
A1 A11 Failed 8/3/2016 8/3/2016
A2 A22 Success 8/2/2016 8/2/2016
A3 A33 Success 8/2/2016 8/2/2016
A4 A44 Failed 8/2/2016 8/2/2016
A4 A44 Failed 8/2/2016 8/2/2016
A4 A44 Success 8/3/2016 8/3/2016
A5 A55 Failed 8/3/2016 8/3/2016
A5 A55 Failed 8/3/2016 8/3/2016
A2 A22 Success 8/3/2016 8/3/2016
我需要输出，其中apname和dbname成功并失败或只有失败状态。我不希望apname和dbname只有状态成功。所有这些条件都应适用于最大日期（例如，基于输入文件的2016年8月3日）
ApName DbName Status StartDate EndDate
A1 A11 Success 8/3/2016 8/3/2016
A1 A11 Failed 8/3/2016 8/3/2016
A5 A55 Failed 8/3/2016 8/3/2016
A5 A55 Failed 8/3/2016 8/3/2016
答案 0 :(得分：0)
因此，如果我理解正确，如果在最长日期"Ariel": {
"adult": false,
"backdrop_path": "\/z2QUexmccqrvw1kDMw3R8TxAh5E.jpg",
"belongs_to_collection": {
"id": 8384,
"name": "Proletariat Collection",
"poster_path": null,
"backdrop_path": "\/ibLeWo3X3dVo4rxvn4m3y90ms1I.jpg"
},
"0": {
"Varjoja paratiisissa": {
"adult": false,
"backdrop_path": "\/6YjUX87VtIEuDzanBE6obVxE9V3.jpg",
"belongs_to_collection": {
"id": 8384,
"name": "Proletariat Collection",
"poster_path": null,
"backdrop_path": "\/ibLeWo3X3dVo4rxvn4m3y90ms1I.jpg"
},
"1": {
"Life in Loops (A Megacities RMX)": {
"adult": false,
"backdrop_path": "\/udvB86uyzJ6P9vmB83WfrCbnmnI.jpg",
"belongs_to_collection": null,
"budget": 42000,
"genres": {
"0": {
"id": 99,
"name": "Documentary"
}
}
至少有一个状态，您想要返回给定apname的所有行吗？
如果是这样，这应该会让您使用failed：
max over使用select t.*
from yourtable t join (
select apname, enddate, status, max(enddate) over () maxenddate
from yourtable
) t2 on t2.apname = t.apname and
t2.maxenddate = t.enddate and
t2.status = 'Failed'
的替代方法：
exists答案 1 :(得分：0)
假设状态为“成功”或“失败”，它应该完成工作：
SELECT * FROM
(
SELECT a.* ,
count(distinct Status) over (partition by a.AppName, trunc(a.EndDate)) as statusCnt,
max(endDate) over(partition by a.AppName) as lastDate
FROM your_table a
)b
WHERE trunc(b.lastDate) = trunc(b.EndDate) AND (statusCnt > 1 OR Status = 'Failed');
我可能会误解有关“最长日期”的部分。我的查询比较了每个AppName的最大日期，而不是整体最大日期。
答案 2 :(得分：0)
这是一个使用分析函数但没有连接的解决方案。
<强>假设：
日期可能包含时间元素（时间）。要求是返回至少有一行具有FAILED状态和EndDate的应用程序/数据库的行，这些行与表中所有enddates的最大值相同（但可能是较早的时间段）。
对于这些应用程序/数据库对，返回表中的所有行。 （如果要求只返回最大日期的行，则问题会更容易。）
<强>查询：
with
inputs ( ApName, DbName, Status, StartDate, EndDate ) as (
select 'A1', 'A11', 'Success', to_date('8/3/2016', 'mm/dd/yyyy'), to_date('8/3/2016', 'mm/dd/yyyy') from dual union all
select 'A1', 'A11', 'Failed' , to_date('8/3/2016', 'mm/dd/yyyy'), to_date('8/3/2016', 'mm/dd/yyyy') from dual union all
select 'A2', 'A22', 'Success', to_date('8/2/2016', 'mm/dd/yyyy'), to_date('8/2/2016', 'mm/dd/yyyy') from dual union all
select 'A3', 'A33', 'Success', to_date('8/2/2016', 'mm/dd/yyyy'), to_date('8/2/2016', 'mm/dd/yyyy') from dual union all
select 'A4', 'A44', 'Failed' , to_date('8/2/2016', 'mm/dd/yyyy'), to_date('8/2/2016', 'mm/dd/yyyy') from dual union all
select 'A4', 'A44', 'Failed' , to_date('8/2/2016', 'mm/dd/yyyy'), to_date('8/2/2016', 'mm/dd/yyyy') from dual union all
select 'A4', 'A44', 'Success', to_date('8/3/2016', 'mm/dd/yyyy'), to_date('8/3/2016', 'mm/dd/yyyy') from dual union all
select 'A5', 'A55', 'Failed' , to_date('8/3/2016', 'mm/dd/yyyy'), to_date('8/3/2016', 'mm/dd/yyyy') from dual union all
select 'A5', 'A55', 'Failed' , to_date('8/3/2016', 'mm/dd/yyyy'), to_date('8/3/2016', 'mm/dd/yyyy') from dual union all
select 'A2', 'A22', 'Success', to_date('8/3/2016', 'mm/dd/yyyy'), to_date('8/3/2016', 'mm/dd/yyyy') from dual
),
prep ( apname, dbname, status, startdate, enddate, fdate, mdate ) as (
select apname, dbname, status, startdate, enddate,
max(case when status = 'Failed' then enddate end) over
(partition by apname, dbname),
max(enddate) over ()
from inputs
)
select apname, dbname, status, startdate, enddate
from prep
where fdate >= trunc(mdate)
;
<强>输出：
APNAME DBNAME STATUS STARTDATE ENDDATE
------ ------ ------- ---------- ----------
A1 A11 Failed 2016-08-03 2016-08-03
A1 A11 Success 2016-08-03 2016-08-03
A5 A55 Failed 2016-08-03 2016-08-03
A5 A55 Failed 2016-08-03 2016-08-03
4 rows selected.