Question

DECLARE @return_value table(BonusSalary int, testid int)

INSERT INTO @return_value 
SELECT ( 
        SUM(case when BonusSalary.Type=1 and BonusSalary.Active=1 then BonusSalary.BonusSalary else 0 end) 
        - SUM(case when BonusSalary.Type=0 and BonusSalary.Active=1 then BonusSalary.BonusSalary else 0 end)
    ) AS Expr1, EmpID
FROM BonusSalary
GROUP BY EmpID 

SELECT  Emp.ID, Emp.EmpName, Emp.NID, EmpEmail.Email, EmpAddress.Address, EmpPhone.Phone, EmpPhone.PhoneType, Salary.Salary, 
        Salary.SalaryType, Positions.PositionName, MainCompany.Name, WorkTimes.StartTime, WorkTimes.EndTime, Contracts.Contract, 
        Contracts.Type, Emp_Contracts.ContractStartDate,@return_value.BonusSalary --, ContractStartEnd.Length, ContractStartEnd.Enddate, ContractStartEnd.Startdate
FROM Emp 
left outer join EmpPhone 
        on Emp.ID=EmpPhone.EmpID
left outer join EmpEmail 
        on Emp.ID=EmpEmail.EmpID
left outer join EmpAddress 
        on Emp.ID=EmpAddress.EmpID
left outer join Emp_Salary 
        ON Emp.ID = Emp_Salary.EmpID 
left outer join Salary 
        ON Emp_Salary.SalaryID = Salary.ID
left outer join Emp_Positions_Company 
        ON Emp.ID = Emp_Positions_Company.EMPID 
left outer join MainCompany 
        ON MainCompany.ID = Emp_Positions_Company.CompanyID 
left outer join Positions 
        on Positions.ID=Emp_Positions_Company.PositionID
left outer join WorkTimes 
        on Emp.ID=WorkTimes.EmpID
left outer join Emp_Contracts 
        on Emp.ID=Emp_Contracts.EmpID 
left outer join Contracts 
        on Contracts.ID=Emp_Contracts.ContractID 
LEFT OUTER JOIN @return_value 
        on Emp.ID = @return_value.testid

我收到了错误消息 Msg 137，Level 16，State 1，Line 14 必须声明标量变量“@return_value”。

Answer 1

如果不使用别名，则无法连接表变量。试试这个：

 SELECT ... ,rv.BonusSalary,...
 ....
 LEFT OUTER JOIN @return_value as rv 
    on Emp.ID = rv.testid

如何在程序中创建表并在select with join中使用？

1 个答案: