如何让 ContactCd 的这个额外聚合在我更大的查询中正常工作？

Question

编辑我最初的问题包括示例数据、我当前的查询和我试图合并的聚合，以便每个 ContactCd 不会创建新行。 我不知道如何使我的示例不那么复杂。 如果我将其简化，是否会删除查询所需的所有连接？

   SELECT        
jobNew_HeaderFile.JobCd, 
GRP = max( case when jobCtrl_Rollup.RollupCatCd = 'GRP' then jobCtrl_Rollup.RollupCd end ), 
REG = max( case when jobCtrl_Rollup.RollupCatCd = 'REG' then jobCtrl_Rollup.RollupCd end ), 
' ' AS 'JOB_1', 
' ' AS 'FIELD OPS MANAGER', 
' ' AS 'NATIONAL OPS MANAGER', 
DIV = max( case when jobCtrl_Rollup.RollupCatCd = 'DIV' then jobCtrl_Rollup.RollupCd end ), 
jobCtrl_Company.Company, 
jobNew_HeaderFile.JobName AS LOCATION, 
jobNew_SiteDetail.JobAddr1 AS Addr1, 
jobNew_SiteDetail.JobAddr2 AS Addr2, 
jobNew_SiteDetail.JobCity AS City, 
jobNew_SiteDetail.JobState AS State, 
jobNew_SiteDetail.JobZip1 AS Zip, 
jobNew_SiteDetail_Contacts.ContactCd,
jobNew_SiteDetail_Contacts.FirstName AS FNAME1, 
jobNew_SiteDetail_Contacts.LastName AS LNAME1, 
jobNew_SiteDetail_Contacts.Title AS TITLE1, 
jobNew_SiteDetail_Contacts.FirstName AS FNAME2, 
jobNew_SiteDetail_Contacts.LastName AS LNAME2, 
jobNew_SiteDetail_Contacts.Title AS TITLE2,
jobNew_SiteDetail_Contacts.FirstName AS FNAME3, 
jobNew_SiteDetail_Contacts.LastName AS LNAME3, 
jobNew_SiteDetail_Contacts.Title AS TITLE3,
jobNew_SiteDetail_Contacts.Phone_wk, 
jobNew_SiteDetail_Contacts.Phone_cell, 
jobNew_SiteDetail_Contacts.email AS Email, 
jobNew_SiteDetail.ICCPhone AS ICC_Phone, 
jobNew_SiteDetail.ICCPhoneNote AS ICC_Phone_Note, 
jobNew_SiteDetail.Email AS ICC_Email, 
mstClients.ClientCd, 
mstClients.ClientName, 
jobCtrl_Rollup.RollupCatCd, 
jobCtrl_Rollup.RollupCd

FROM jobNew_HeaderFile 
INNER JOIN jobNew_SiteDetail ON jobNew_HeaderFile.JobCd = jobNew_SiteDetail.JobCd 
INNER JOIN jobNew_SiteDetail_Contacts ON jobNew_HeaderFile.JobCd = jobNew_SiteDetail_Contacts.JobCd 
INNER JOIN jobCtrl_Company ON jobNew_HeaderFile.JobCd = jobCtrl_Company.JobCd 
INNER JOIN mstClients ON jobNew_SiteDetail.ClientCd = mstClients.ClientCd 
INNER JOIN jobCtrl_Rollup ON jobNew_HeaderFile.JobCd = jobCtrl_Rollup.JobCd

WHERE 
jobNew_HeaderFile.JobCd = 990
--AND jobNew_SiteDetail.ActiveStatus = '1'
AND jobCtrl_Rollup.CurrentYN = '1'

GROUP BY
jobNew_HeaderFile.JobCd, 
jobCtrl_Company.Company, 
jobNew_HeaderFile.JobName, 
jobNew_SiteDetail.JobAddr1, 
jobNew_SiteDetail.JobAddr2, 
jobNew_SiteDetail.JobCity, 
jobNew_SiteDetail.JobState, 
jobNew_SiteDetail.JobZip1, 
jobNew_SiteDetail_Contacts.ContactCd,
jobNew_SiteDetail_Contacts.FirstName, 
jobNew_SiteDetail_Contacts.LastName, 
jobNew_SiteDetail_Contacts.Title, 
jobNew_SiteDetail_Contacts.Phone_wk, 
jobNew_SiteDetail_Contacts.Phone_cell, 
jobNew_SiteDetail_Contacts.email, 
jobNew_SiteDetail.ICCPhone, 
jobNew_SiteDetail.ICCPhoneNote, 
jobNew_SiteDetail.Email, 
mstClients.ClientCd, 
mstClients.ClientName, 
jobCtrl_Rollup.RollupCatCd, 
jobCtrl_Rollup.RollupCd

我试图包含的汇总是：

Select JobCd
      ,FName1 = max( case when RN=1 then FName end)
      ,LName1 = max( case when RN=1 then LName end)
      ,FName2 = max( case when RN=2 then FName end)
      ,LName2 = max( case when RN=2 then LName end)
      ,FName3 = max( case when RN=3 then FName end)
      ,LName3 = max( case when RN=3 then LName end)
 From ( 
        Select JobCd
              ,RN    = row_number() over (partition by JobCd order by [ContactCd])
              ,FName = parsename(replace([FirstName],' ','.'),1)
              ,LName = parsename(replace([LastName],' ','.'),1)
         From  jobNew_SiteDetail_Contacts
      ) A

     Group By JobCd

想要的结果：

工作CD	F名称1	L名称1	F名称2	L名称2	F名称3	L名称3
990	贝基	史密斯	杰弗里	梅森	基督教	福腾伯格

Answer 1

这是无法回答的，比如我们不知道Job表和其他表之间有什么关系。是1-多等吗

但是为什么不先进行所需的聚合，然后再次连接其余表呢？这会简化您的查询。

类似：

select * from (
   select ID, MAX(...) from
   Sometable GROUP BY ID
  ) x
INNER JOIN SomeDetails sd
   ON sd.ID = x.ID 
INNER JOIN SomeMoreDetails sd2 ON sd2.ID = x.ID`
... rince and repeat

这样您就不需要聚合十亿个字段。

此外，您的查询示例有：

jobNew_HeaderFile.JobName AS LOCATION,  jobNew_SiteDetail.JobAddr1 AS Addr1,  jobNew_SiteDetail.JobAddr2 AS Addr2,  jobNew_SiteDetail.JobCity AS City,  jobNew_SiteDetail.JobState AS State,  jobNew_SiteDetail.JobZip1 AS Zip, 

您认为您必须指定问题中的所有列吗？在我看来，如果你的问题只是：

jobNew_HeaderFile.JobName AS LOCATION