Question

我的问题如下。我希望能够使所有在等级上低于给定员工（假设ID为73）的员工，并计算这些员工与他之间的等级级别数。但是，似乎没有任何工作适合我。到目前为止，我已经尝试了这段代码。但是，它似乎仅在id在层次结构中上升时变小时才起作用，这不是这种情况（并且它不计算深度）

select  EmployeeKey,
        Employeecol 
from    (select * from Employee
         order by Employeecol, EmployeeKey) Employee_sorted,
        (select @pv := '73') initialisation
where   find_in_set(Employeecol, @pv)
and     length(@pv := concat(@pv, ',', EmployeeKey))

表格：

CREATE TABLE [Employee](
    [EmployeeKey] [int] NOT NULL,
    [ParentEmployeeKey] [int] NULL,
    [SalesTerritoryKey] [int] NULL,
    [FirstName] [nvarchar](50) NOT NULL,
    [LastName] [nvarchar](50) NOT NULL,
    [HireDate] [date] NULL,
    [Gender] [nchar](1) NULL
)

数据：

1,18,11,"Guy","Gilbert",2006-01-28,"M"
2,7,11,"Kevin","Brown",2006-08-26,"M"
3,14,11,"Roberto","Tamburello",2007-06-11,"M"
4,3,11,"Rob","Walters",2007-07-05,"M"
5,3,11,"Rob","Walters",2007-07-05,"M"
6,267,11,"Thierry","D'Hers",2007-07-11,"M"
7,112,11,"David","Bradley",2007-07-20,"M"

谢谢您的反馈

MySQL获得所有子项和深度

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