Question

你好，我被困在这里一个星期...... 说我在查询中有这个结果..

Branch   Pay1   Pay2   Pay3   Pay4
Branch1  100    100    100    100
Branch1  150    150    150    150
Branch2  200    200    200    200
Branch3  200    200    200    200

我想做这样的事情

Branch   Pay1   Pay2   Pay3   Pay4
Branch1  250    250    250    250
Branch2  200    200    200    200
Branch3  200    200    200    200

最后的结果是这样的

Item   Branch1   Branch2   Branch3   
pay1   250       200       200
pay2   250       200       200   
pay3   250       200       200    
pay4   250       200       200

希望你能帮助我做这件事......非常感谢..

这是我的查询...第一个结果

Select distinct 
    --pr_employees.Fullname as Name
   --,
   PR_Employees.BranchID,PR_payroll.BasicPay as [BasicPay]
   ,PR_Empearnings.EarningAmt
   ,PR_Earnings.Description
   ,pr_payroll.Overtime
   ,pr_payroll.Period
   ,Pr_payroll.SundayOT as [Sunday OT]
   ,Pr_Payroll.PaidHol as [Paid Hol]
   ,pr_payroll.ThirteenthMonthPay as [Thirteen MO]
   ,pr_payroll.Grosspay as [Gross Amount]
   ,pr_payroll.WithHoldingTax as [WithTax]
   ,pr_payroll.SSSPremium as [SSS Cont]
   ,pr_payroll.SSSLoan as [SSS Loan]
   ,pr_payroll.PagibigPremium as [Pagibig Cont]
   ,pr_payroll.PagibigLoan as [Pagibig Loan]
   ,pr_payroll.NHIPPremium as Medicare
   ,pr_payroll.TotalDeductions as [Total Ded]
   ,pr_Payroll.netpay as [Net with OD]
   ,pr_payroll.netnoOd as [Net no OD]
   ,prchargesAdvances.Credit
   ,prchargesadvancesTypes.ChargesTypeName
 from pr_employees
  left join pr_payroll on PR_Employees.EmpID=PR_Payroll.EmpID
  left join PR_EmpEarnings on PR_Payroll.EmpID=PR_EmpEarnings.EmpID
  left join PR_Earnings on PR_EmpEarnings.EarningId=pr_earnings.earningid
  left join PR_Overtime on PR_Overtime.EmpID=PR_Payroll.EmpID
  left join PRChargesAdvances on PRChargesAdvances.transactiondate=pr_payroll.period   
   and prchargesadvances.empid=pr_payroll.empid
  left join PRChargesAdvancesTypes on PRChargesAdvances.ChargeTypeID=PRChargesAdvancesTypes.ChargesTypeID
 where PR_Payroll.Period='8/31/2013'

Answer 1

这不仅仅是一个模板，而是一个解决方案，但我会解释一下。

在下面的with子句中，您将按分支建立聚合。如果您将其作为自己的查询独立运行（重命名并添加所有付费列），这将为您提供所需输出的第一步（您帖子中的第一个所需结果集）。

然后你运行几个由工会连接的查询，你在其中指定你要汇总的薪水栏的文字（你必须把它作为文字输入，这就是为什么我用'literal'这个词来每次都说明。然后总结一下，每个pay字段一行，每个分支列（通过case语句），with子句中查询的聚合。我使用case语句之和的唯一原因是这样你就可以在每个薪资领域的同一行获得所有#s。

您有超过4个付费列（如您所知），因此您必须在完成之前添加几个联盟（并将所有付费列和分支ID重命名为它们）。

with sub as
 (select branch,
         sum(pay1) as sum_py1 sum(pay2) as sum_py2 sum(pay3) as sum_py3 sum(pay4) as sum_py4
    from bunchoftbls
   where PR_Payroll.Period = '8/31/2013'
   group by branch)
select 'Pay1 literal'
        , sum(case when branch = 'Branch1' then sum_py1 else 0 end) as branch1
        , sum(case when branch = 'Branch2' then sum_py1 else 0 end) as branch2
        , sum(case when branch = 'Branch3' then sum_py1 else 0 end) as branch3
        , sum(case when branch = 'Branch4' then sum_py1 else 0 end) as branch4
        , sum(case when branch = 'Branch5' then sum_py1 else 0 end) as branch5
        , sum(case when branch = 'Branch6' then sum_py1 else 0 end) as branch6
from sub
union all
select 'Pay2 literal'
        , sum(case when branch = 'Branch1' then sum_py2 else 0 end) as branch1
        , sum(case when branch = 'Branch2' then sum_py2 else 0 end) as branch2
        , sum(case when branch = 'Branch3' then sum_py2 else 0 end) as branch3
        , sum(case when branch = 'Branch4' then sum_py2 else 0 end) as branch4
        , sum(case when branch = 'Branch5' then sum_py2 else 0 end) as branch5
        , sum(case when branch = 'Branch6' then sum_py2 else 0 end) as branch6
from sub
union all
select 'Pay3 literal'
        , sum(case when branch = 'Branch1' then sum_py3 else 0 end) as branch1
        , sum(case when branch = 'Branch2' then sum_py3 else 0 end) as branch2
        , sum(case when branch = 'Branch3' then sum_py3 else 0 end) as branch3
        , sum(case when branch = 'Branch4' then sum_py3 else 0 end) as branch4
        , sum(case when branch = 'Branch5' then sum_py3 else 0 end) as branch5
        , sum(case when branch = 'Branch6' then sum_py3 else 0 end) as branch6
from sub
union all
select 'Pay4 literal'
        , sum(case when branch = 'Branch1' then sum_py4 else 0 end) as branch1
        , sum(case when branch = 'Branch2' then sum_py4 else 0 end) as branch2
        , sum(case when branch = 'Branch3' then sum_py4 else 0 end) as branch3
        , sum(case when branch = 'Branch4' then sum_py4 else 0 end) as branch4
        , sum(case when branch = 'Branch5' then sum_py4 else 0 end) as branch5
        , sum(case when branch = 'Branch6' then sum_py4 else 0 end) as branch6
from sub

用于报告的Sql枢轴

1 个答案: