1到多个具有聚合函数的Select查询(Adventure Works数据库)

问题描述 投票:0回答:1

我正在使用Adventure Works数据库练习SQL。我的任务很简单。我想找一个员工,看看他们赚多少钱。

名字|姓氏|年龄|支付率|

问题是,工资率位于表中,该表与ModifiedDate列与雇员(EmployeePayHistory)的一对多关系。我想获取最新的ModifiedDate,但没有尝试过的方法。我一直在子查询中追赶聚合函数


SELECT e.BusinessEntityID,p.FirstName [First Name], p.LastName [Last Name], DATEDIFF(YEAR,e.BirthDate, GETDATE() )[Age],
(SELECT eph1.Rate FROM HumanResources.EmployeePayHistory eph1 HAVING eph1.Rate = MAX(eph.ModifiedDate))
FROM Person.Person p 
JOIN HumanResources.Employee e ON p.BusinessEntityID = e.BusinessEntityID
JOIN HumanResources.EmployeePayHistory eph ON e.BusinessEntityID = eph.BusinessEntityID
GROUP BY e.BusinessEntityID, p.FirstName,p.LastName, DATEDIFF(YEAR,e.BirthDate, GETDATE() )

sql sql-server tsql greatest-n-per-group adventureworks
1个回答
0
投票

这是每组最大的问题。不要认为聚合:请考虑过滤。

我想这就是您想要的:

SELECT 
    e.BusinessEntityID
    p.FirstName [First Name], 
    p.LastName [Last Name], 
    DATEDIFF(YEAR,e.BirthDate, GETDATE()) [Age],
    (
        SELECT TOP (1) eph.Rate 
        FROM HumanResources.EmployeePayHistory eph 
        WHERE eph.BusinessEntityID = p.BusinessEntityID
        ORDER BY eph.ModifiedDate DESC
    ) [LatestRate]
FROM Person.Person p 
JOIN HumanResources.Employee e ON p.BusinessEntityID = e.BusinessEntityID

注意:您没有确切说明PersonEmployee之间的关系-上面假设这是1-1关系。

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