Question

我有一个名为Issue_Status的列，它显示了主要与风险管理相关的问题的当前状态。我需要在Cognos和/或SQL Server中创建一个新列，以显示Issue的所有先前状态的连接列。它是这样的。

现在，我没有Issue_Status_Concatenated。

就是这样。

Issue I.D   Issue Status   

1234        Draft           
1234        Open           
1234        Closed          
5678        Draft           
5678        Pending         
5678        Closed          
5678        Cancelled       
3333        Draft           
3333        Pending         
3333        Pending         
3333        Pending

我在考虑使用运行计数功能（Cognos Function）

Issue I.D   Issue Status         Running_Count

1234        Draft                1
1234        Open                 2
1234        Closed               3
5678        Draft                1
5678        Pending              2
5678        Closed               3
5678        Cancelled            4
3333        Draft                1
3333        Pending              2
3333        Pending              3
3333        Pending              4

然后使用案例函数

将结果合并到单个列中

New Concat_Column 1

case when Running_Count = 1 then Issue_status else ' ' end

New Concat_Column 2 

case when Running_Count = 2 then Issue_status else ' ' end

New Concat_Column 3 

case when Running_Count = 3 then Issue_status else ' ' end

New Concat_Column 4

case when Running_Count = 4 then Issue_status else ' ' end

然后我计划通过将所有New concat列从1组合到

创建Issue_Status_Concatenated来创建新数据项或新列

New Concat_Column 1 + '|' + New Concat_Column  2 + '|' + New Concat_Column 3 + '|' + New Concat_Column 4

我知道这是一个漫长的过程，但我知道可以采用更简单合理的方式来实现这一目标吗？有没有办法让这更简单？

Issue I.D   Issue Status   Issue_Status_Concatenated

    1234        Draft           Draft | Open | Closed
    1234        Open            Draft | Open | Closed
    1234        Closed          Draft | Open | Closed
    5678        Draft           Draft | Open | Closed |Cancelled
    5678        Pending         Draft | Open | Closed |Cancelled
    5678        Closed          Draft | Open | Closed |Cancelled
    5678        Cancelled       Draft | Open | Closed |Cancelled
    3333        Draft           Draft | Pending
    3333        Pending         Draft | Pending
    3333        Pending         Draft | Pending
    3333        Pending         Draft | Pending

Answer 1

下面的片段将帮助您在oracle中获取deird输出。希望这会有所帮助。

SELECT ID,
  STAT,
  LISTAGG(STAT,'|') WITHIN GROUP(
ORDER BY STAT) OVER(PARTITION BY ID) AGG_STAT
FROM
  (WITH TMP AS
  ( SELECT 1234 id, 'Draft' Stat FROM dual
  UNION ALL
  SELECT 1234 ID, 'Open' STAT FROM DUAL
  UNION ALL
  SELECT 1234 id, 'Completed' Stat FROM dual
  UNION ALL
  SELECT 1100 ID, 'Draft' STAT FROM DUAL
  UNION ALL
  SELECT 1100 ID, 'Pending' STAT FROM DUAL
  UNION ALL
  SELECT 1100 ID, 'Completed' STAT FROM DUAL
  UNION ALL
  SELECT 1100 id, 'Closed' Stat FROM dual
  )
SELECT tmp.*,ROW_NUMBER() OVER(PARTITION BY ID ORDER BY 1 DESC) RN FROM tmp
  );

Answer 2

如果它的sql server是

SELECT  i.issue_id,
    stuff((
            SELECT DISTINCT ' | ' + issue_status
            FROM issues i1
            WHERE i1.issue_id = i.issue_id for xml path('')),1,3,'')
            FROM issues i

如果它的oracle它的概率沿着listagg的行 https://docs.oracle.com/cd/E11882_01/server.112/e41084/functions089.htm#SQLRF30030

Answer 3

对于SQL Server，您可以使用以下查询来获取结果 -

Create table #IssueLog (
[Issue I.D] int ,
[Issue Status] varchar(10)
)

insert into #IssueLog
select 1234 ,       'Draft'
union all  select 1234,     'Open'           
union all  select 1234,      'Closed'          
union all  select 5678,     'Draft'           
union all  select 5678,      'Pending'         
union all  select 5678,     'Closed'          
union all  select 5678,     'Cancelled'       
union all  select 3333,     'Draft'           
union all  select 3333,     'Pending'         
union all  select 3333,     'Pending'         
union all  select 3333,      'Pending'   


Select Main.[Issue I.D],
       Left(Main.[Issues],Len(Main.[Issues])-1) As [Issues]
From
    (
        Select distinct T2.[Issue I.D] ,
            (
                Select T1.[Issue Status] + '|' AS [text()]
                From #IssueLog T1
                Where T1.[Issue I.D] = T2.[Issue I.D]
                ORDER BY T1.[Issue I.D]
                For XML PATH ('')
            ) [Issues]
        From #IssueLog T2
    ) [Main]

如果从单个列获取数据，如何创建连续列？

3 个答案: