Question

有表Employees

CREATE TABLE Employees
(
    id int NOT NULL IDENTITY(1, 1) PRIMARY KEY,
    name  nvarchar(100) NOT NULL,
    depID int NOT NULL,
    salary money NOT NULL,
    FOREIGN KEY (depID) REFERENCES Departments(id)
);

和Payments

CREATE TABLE Payments
(
    id int NOT NULL IDENTITY(1, 1) PRIMARY KEY,
    userID int NOT NULL,
    createdDate date DEFAULT GETDATE(),
    sum money NOT NULL,
    FOREIGN KEY (userID) REFERENCES Employees(id)
);

过去两年我需要获得前三名薪水的员工姓名我尝试使用下面的查询，但它没有工作，我收到了一个错误。

SELECT TOP 3 name
FROM Employees
WHERE id in (SELECT id, SUM(sum) as SumTotal FROM Payments
             WHERE (createdDate BETWEEN '2015-09-01' AND '2013-09-01')
             ORDER BY SumTotal);

错误讯息：

ORDER BY子句在视图，内联函数，派生表，子查询和公用表表达式中无效，除非还指定了TOP或FOR XML。

如何让它发挥作用？

Answer 1

这是使用cte s。

执行此操作的一种方法

Demo

with pymt as 
(
SELECT userid, sum(sum) as sumtotal
FROM Payments
WHERE createdDate BETWEEN '2013-09-01' AND '2015-09-01'
group by userid
)
, ename as 
(
select e.name, pymt.sumtotal, row_number() over(order by pymt.sumtotal desc) as rn
from pymt join employees e
on pymt.userid = e.id
)
select name 
from ename 
where rn < = 3;

Answer 2

试试这个： -

SELECT userID, SUM(sum) as SumTotal INTO #temp
FROM Payments
WHERE (createdDate BETWEEN '2015-09-01' AND '2013-09-01')
group by userID
ORDER BY SumTotal

SELECT TOP 3 name
FROM Employees e join #temp
on e.id = #temp.userID

Answer 3

您必须执行错误消息告诉您的操作，将TOP放在您的订单所在地：

SELECT name
FROM Employees
WHERE id in (SELECT TOP 3 id, SUM(sum) as SumTotal FROM Payments
             WHERE (createdDate BETWEEN '2015-09-01' AND '2013-09-01')
             ORDER BY SumTotal)

[编辑] 如果我们按照错误消息，这应该没问题：

SELECT name
FROM Employees
WHERE id in ( Select x.userId From (SELECT TOP 3 userId, SUM([sum]) as SumTotal FROM Payments
             WHERE (createdDate BETWEEN '2015-09-01' AND '2013-09-01')
             Group By userId
             ORDER BY SumTotal) x);

Answer 4

select top 3
emp.id as id,
emp.name as name,
sum(pay.sum) as total
from employees emp
join payments pay
    on emp.id = pay.user_id
--where pay.createdDate BETWEEN '2015-09-01' AND '2013-09-01'
where pay.createdDate BETWEEN '2013-09-01' AND '2015-09-01'
group by emp.id, emp.name

这应该有效

Answer 5

人们过于复杂化了。我很确定这会得到你想要的东西：

SELECT TOP 3 employees.id, Name,
             Sum([sum]) AS [TotalPayments]
FROM Employees
inner join Payments on employees.id = payments.userid
WHERE createdDate BETWEEN  '2013-09-01' and '2015-09-01'
Group By employees.id, Name
Order BY TotalPayments DESC

SqlFiddle来测试它

如果只需要names列，可以使用另一个select包装该查询：

select Name from (
  SELECT TOP 3 employees.id, Name,
    Sum([sum]) AS [TotalPayments]
  FROM Employees
  inner join Payments on employees.id = payments.userid
  WHERE createdDate BETWEEN  '2013-09-01' and '2015-09-01'
  Group By employees.id, Name
  Order BY TotalPayments DESC
) q

Answer 6

SELECT TOP 3 Name,
             Sum(sum) AS sum
FROM Employees
WHERE createdDate BETWEEN '2015-09-01' AND '2013-09-01'
Group By Name
Order BY 2 DESC

选择过去两年中前三名薪水的员工

6 个答案: