我有一张桌子[dbo]。[RISK_DISCPLN]
RISK_DISCPLN_ID RISK_DISCPLN_NM
1 LegalRisk
2 institutional Risk
4 Market Risk
Tabbe名称[dbo]。[PROJ_RISK_DISCPLN]
PROJ_ID RISK_DISCPLN_ID
1 1
1 2
1 4
2 1
2 2
2 4
3 1
3 2
3 4
表名[dbo]。[USER]
USER_ID USER_FIRST_NM USER_LST_NM
2 saravanakumar rajkumar
3 Soosai Antony
4 Adam Allen
5 Babita Tripathy
9 stacey Davis
11 NULL NULL
我尝试在下面查询显示相应风险的名称,我得到所有的名称为空值...请帮助
;with cte as
(
SELECT [PROJ_RISK_DISCPLN].PROJ_ID,
[USER].USER_LST_NM + ' '+ [USER].USER_FIRST_NM as Name,
[RISK_DISCPLN].RISK_DISCPLN_NM,
[RISK_DISCPLN].RISK_DISCPLN_ID,
[USER].[USER_ID]
FROM dbo.[PROJ_RISK_DISCPLN]
left join [dbo].[USER]
on [PROJ_RISK_DISCPLN].RISK_OWN_USER_ID = [dbo].[USER].[USER_ID]
left join dbo.[RISK_DISCPLN] on
[PROJ_RISK_DISCPLN].RISK_DISCPLN_ID = [RISK_DISCPLN].RISK_DISCPLN_ID
)
select *
from
(
select c1.PROJ_ID,
c1.RISK_DISCPLN_NM,
STUFF(
(SELECT ', ' + c2.Name
FROM cte c2
where c1.PROJ_ID = c2.PROJ_ID
and c1.RISK_DISCPLN_ID = c2.RISK_DISCPLN_ID
FOR XML PATH (''))
, 1, 1, '') AS Name
from cte c1
) d
pivot
(
max(Name)
for RISK_DISCPLN_NM in ([LegalRisk Owner],[institutional Risk Owner],[Market Risk Owner])
) piv
答案 0 :(得分:0)
我得到了回答,而是[LegalRisk所有者],[机构风险所有者],[市场风险所有者] 我必须使用[LegalRisk],[机构风险],[市场风险] ....