Question

不要评判我，我是SQL查询的新手。我得到了计划，如下图所示。因此，有2个表，第一个Employees包含EmployeeID，FirstName，LastName，DateOfBirth和DepartmentID。第二个叫做Department，包含DepartmentID和DepartmentName。

我想为每个包含20名以上员工的部门中最老的员工返回FirstName，LastName和DepartmentName。

我的解决方案是以下查询：

SELECT FirstName, LastName, DepartmentName
FROM employees
LEFT JOIN department
ON employees.DepartmentID = department.DepartmentID
    WHERE (employees.DateOfBirth = 
(SELECT MIN(employees.DateOfBirth ) FROM (
SELECT *FROM employees WHERE employees.DepartmentID IN ( 
SELECT employees.DepartmentID FROM employees GROUP BY DepartmentID HAVING COUNT(*) > 20)));

我认为逻辑很好，因为内部SELECT语句将返回每个拥有超过20名员工的部门的ID，而外部应该返回给最老的员工。

我遇到的问题是当我尝试执行此查询时，它返回SQL error每个派生表必须拥有它自己的别名。我尝试在每个派生表上添加别名，但结果仍然相同。请帮帮我。

此外，如果有人有不同的解决方案，请分享。

谢谢。添加哪个草莓要求，Create查询

CREATE TABLE Employees
(
EmployeeID int, 
FirstName varchar(10),
LastName varchar(15),
DateOfBirth date,
DeparmentID int
)

CREATE TABLE Department
(
DepartmentID int, 
DepartmentName varchar(15)
)

Answer 1

由于格式不一致，您的查询很难阅读。所以我会按照以下方式清理它：

SELECT  FirstName, LastName, DepartmentName
FROM    employees
        LEFT JOIN department
        ON employees.DepartmentID = department.DepartmentID
WHERE   (employees.DateOfBirth = 
        (
        SELECT  MIN(employees.DateOfBirth)
        FROM    (
                SELECT  *
                FROM    employees
                WHERE   employees.DepartmentID IN (
                        --Departments with more than 20 employees
                        SELECT  employees.DepartmentID
                        FROM    employees
                        GROUP BY DepartmentID
                        HAVING COUNT(*) > 20)
                ) -- You need an alias here.
                  -- Also from this point you were missing closing brackets.

查询问题：

显然，丢失的别名和右括号意味着您甚至无法测试您的查询。
SELECT MIN(employees.DateOfBirth)也只返回一个值。不是每个部门的价值。
因此，您的整体结果仅包括所有“大”部门中最年长的员工。（除非每个部门中最老的员工碰巧有相同的出生日期。）
如果任何员工的出生日期与大部门中最早的出生日期相同，那么它还可能包括较小部门的结果。那个员工甚至不需要成为他们部门中最老的员工！
使用比必要更多的子查询也会导致效率低下。

CTE（公用表表达式）非常适合简化复杂查询。但我不知道mysql是否支持它们。所以这个解决方案仍然使用子查询。

SELECT  e.FirstName, e.LastName, d.DepartmentName
FROM    employees e -- I prefer short aliases
        INNER JOIN (
        -- This sub-query returns the earliest birth date within each
        -- big department. This needs to be an aliased query so you
        -- can join to other tables for your desired columns.
        SELECT  DepartmentID, MIN(DateOfBirth) AS MinDOB -- Must alias column
        FROM    employees
        WHERE   DepartmentID IN (
                -- Big departments
                SELECT  DepartmentID
                FROM    employees
                GROUP BY DepartmentID
                HAVING COUNT(*) > 20
                )
        GROUP BY DepartmentID
        ) ddob -- Alias Department Date of Birth
        -- As a result of inner joining to ddob your employees
        -- will be filtered to only those that match the relevant
        -- ones identified in the query.
        ON  e.DepartmentID = ddob.DepartmentID
        AND e.DateOfBirth = ddob.MinDOB
        INNER JOIN Department d
        ON  d.DepartmentID = e.DepartmentID

在上述解决方案中需要注意的一点是，如果2名员工因部门中最老的而被捆绑，则两者都将被退回。

这种方法在结构上与您的方法类似，但您也可以从另一个方向处理问题。

开始让所有部门的老员工。
最后只根据部门规模过滤结果。

我会留给你试试。我怀疑查询会更简单一些。

Answer 2

以下SQL脚本对应于您的类图：

CREATE TABLE Departments (
 DepartmentID int AUTO_INCREMENT PRIMARY KEY, 
 DepartmentName varchar(15)
);

CREATE TABLE Employees (
 EmployeeID int AUTO_INCREMENT PRIMARY KEY, 
 FirstName varchar(10),
 LastName varchar(15),
 DateOfBirth date,
 DepartmentID int,
 FOREIGN KEY (DepartmentID) REFERENCES Departments(DepartmentID)
);

按照您的班级图表，每个员工都有一个部门。那么，是使用INNER JOIN的原因。我认为以下查询可以执行您想要的操作：

SELECT ee.FirstName, ee.LastName, ee.DateOfBirth, t.DepartmentName 
FROM
( 
     SELECT e.DepartmentID, d.DepartmentName, MIN(e.DateOfBirth) AS DateOfBirth 
     FROM Employees AS e
     INNER JOIN Departments AS d ON e.DepartmentID = d.DepartmentID
     WHERE e.DepartmentID IN (
         SELECT DepartmentID 
         FROM Employees 
         GROUP BY DepartmentID HAVING COUNT(DepartmentID) > 20
     )
     GROUP BY e.DepartmentID
) AS t
INNER JOIN Employees AS ee
WHERE ee.DepartmentID = t.DepartmentID AND ee.DateOfBirth = t.DateOfBirth

输出示例：

FirstName   LastName    DateOfBirth DepartmentName
fisrt14     last14      02/01/2000  SI
fisrt31     last31      12/01/2003  Finance

你改善了它的表现！

如果部门员工超过20人，则返回有关部门中最老员工的信息

2 个答案: