MySQL的高级加入

Question

我最近一直在学习MySQL和想知道如果有人可以帮助我解决这个问题我的工作..

我有两个表，

首先是departmentTable

+------------+-----------+-----------+-----------+-----------+-----------+-----------+ 
| Department | Employee1 | Employee2 | Employee3 | Employee4 | Employee5 | Employee6 | 
+------------+-----------+-----------+-----------+-----------+-----------+-----------+ 
| Finance | Tim C  | Nora C | Susan P | Martin S | Donal D | Zita P | 
| HR   | Yvonne S | Marcus F | Mark B | Peter S | Ola G  | Gina P | 
| IT   | John W | Matt R | Paul M | Tom Y  | Jean S | Martin T | 
+------------+-----------+-----------+-----------+-----------+-----------+-----------+ 

二是employeeDetails表

+--------------+-------+------------+--------+ 
| EmployeeName | Shift | Employment | Salary | 
+--------------+-------+------------+--------+ 
| Donal D  | Night | Full  | 30000 | 
| Gina P  | Night | Part  | 21000 | 
| Jean S  | Night | Full  | 30000 | 
| John W  | Night | Full  | 24000 | 
| Marcus F  | Night | Full  | 21000 | 
| Mark B  | Day | Full  | 23000 | 
| Martin S  | Day | Full  | 23400 | 
| Martin T  | Day | Part  | 31000 | 
| Matt R  | Day | Full  | 22000 | 
| Nora C  | Day | Full  | 22000 | 
| Ola G  | Night | Full  | 21400 | 
| Paul M  | Day | Part  | 23000 | 
| Peter S  | Day | Part  | 25000 | 
| Susan P  | Day | Part  | 23000 | 
| Tim C  | Day | Full  | 24000 | 
| Tom Y  | Day | Full  | 23400 | 
| Yvonne S  | Day | Full  | 24000 | 
| Zita P  | Night | Part  | 31000 | 
+--------------+-------+------------+--------+ 

我试图找到一种方法如果有可能查询哪个部门有4天和2夜班员工。所以它应该输出是这样的..

+------------+-----------+-----------+-----------+-----------+-----------+-----------+ 
| Department | Employee1 | Employee2 | Employee3 | Employee4 | Employee5 | Employee6 | 
+------------+-----------+-----------+-----------+-----------+-----------+-----------+ 
| Finance | Tim C  | Nora C | Susan P | Martin S | Donal D | Zita P | 
| IT   | John W | Matt R | Paul M | Tom Y  | Jean S | Martin T | 
+------------+-----------+-----------+-----------+-----------+-----------+-----------+ 

任何帮助表示赞赏，我知道这是不是存储数据并获取他们最好的方式，但我想了解和学习更多关于MySQL的加入和计数。

Answer 1

你应该通过在一列中存储员工来修复你的表结构。

虽这么说，你可以试试这个：

select 
    d.* 
from departmentTable d 
join employeeDetails e on e.EmployeeName in (
    d.Employee1, d.Employee2, d.Employee3, 
    d.Employee4, d.Employee5, d.Employee6 
) 
group by d.department 
having sum(e.shift = 'Night') = 2 
and sum(e.shift = 'Day') = 4 

Answer 2

您应该使用一张表格，您可以在其中存储所有员工的详细信息并在其中添加引用某个部门的详细信息。不要在employee1-n中使用你的方法。

另外，当我看到Nora C出现在两个部门时，可能会看看m-n关系。 Google是你的朋友