Full recursive employee-boss relation in SQL Server - sql

Question

I need to get the name of all of the employees that depends of a person directly or indirectly. Using the query in this example (from https://rextester.com/WGVRGJ67798),
create table employee(
id int not null,
employee varchar(10) not null,
boss int null
)
insert into employee values
(1,'Anna',null),
(2,'Bob',1),
(3,'Louis',1),
(4,'Sara',2),
(5,'Sophie',2),
(6,'John',4);
with boss as (
select id, employee, boss, cast(null as varchar(10)) as name
from employee
where boss is null
union all
select e.id, e.employee, b.id, b.employee
from employee e
join boss b on b.id = e.boss
)
select * from boss
I can get this result:
However, I need to see this:
It would be like showing all the possible relations between a person an all of those employees "below" him or her.

Answer 1

You can reverse the logic: instead of starting from the boss (the root) and going towards employees (the leafs), you could start from the leafs and walk toward the root. This lets you generate the intermediate relations as you go:
with cte as (
select e.id, e.employee, e.boss, b.employee name, b.boss new_boss
from employee e
left join employee b on b.id = e.boss
union all
select c.id, c.employee, c.new_boss, e.employee, e.boss
from cte c
join employee e on e.id = c.new_boss
)
select id, employee, boss, name
from cte
order by id, boss
Demo on DB Fiddle:
id | employee | boss | name
-: | :------- | ---: | :---
1 | Anna | null | null
2 | Bob | 1 | Anna
3 | Louis | 1 | Anna
4 | Sara | 1 | Anna
4 | Sara | 2 | Bob
5 | Sophie | 1 | Anna
5 | Sophie | 2 | Bob
6 | John | 1 | Anna
6 | John | 2 | Bob
6 | John | 4 | Sara

Answer 2

I like hierarchyid for this sort of thing.
use tempdb;
drop table if exists employee;
drop table if exists #e;
create table employee(
id int not null,
employee varchar(10) not null,
boss int null
)
insert into employee values
(1,'Anna',null),
(2,'Bob',1),
(3,'Louis',1),
(4,'Sara',2),
(5,'Sophie',2),
(6,'John',4);
with boss as (
select id, employee, boss,
cast(concat('/', id, '/') as hierarchyid) as h
from employee
where boss is null
union all
select e.id, e.employee, b.id,
cast(concat(b.h.ToString(), e.id, '/') as hierarchyid)
from employee e
join boss b on b.id = e.boss
)
select *
into #e
from boss
select e.id, e.employee, b.id, b.employee, b.h.ToString()
from #e as e
left join #e as b
on e.h.IsDescendantOf(b.h) = 1
and e.id <> b.id;
I took your code mostly as is and changed the following things:
Rather than keeping track of the boss in the recursive CTE, I'm building a hierarchyid path that leads all the way back to the root of the hierarchy.
Shoved the results of the cte into a temp table
Selected from the temp table, using a self-join where the join criteria are "where the inner table's notion of employee is anywhere in the management chain for the outer table".
Note, for the join, I'm excluding the case where the employee reports to themselves; you cannot be your own boss in this situation (even though the IsDescendantOf method would suggest otherwise!).

Answer 3

Something like this. There are two recursions. First, to get the h_level which with the first recursion represent boss-->employee relationships. Second, treats each row from the first as the leaf node in a new recursion to find direct and indirect hierarchical relationships.
Data
drop table if exists Employee;
go
create table employee(
id int not null,
employee varchar(10) not null,
boss int null)
insert into employee values
(1,'Anna',null),
(2,'Bob',1),
(3,'Louis',1),
(4,'Sara',2),
(5,'Sophie',2),
(6,'John',4);
Query
;with
boss(id, employee, boss, h_level) as (
select id, employee, boss, 0
from employee
where boss is null
union all
select e.id, e.employee, b.id, b.h_level+1
from employee e
join boss b on b.id = e.boss),
downlines(id, employee, boss, h_level, d_level) as (
select id, employee, boss, h_level, 0
from boss
union all
select b.id, b.employee, d.id, d.h_level, d.d_level+1
from boss b
join downlines d on d.id = b.boss)
select *
from downlines
order by h_level, d_level;
Output
id employee boss h_level d_level
1 Anna NULL0 0
2 Bob 1 0 1
3 Louis 1 0 1
4 Sara 2 0 2
5 Sophie 2 0 2
6 John 4 0 3
2 Bob 1 1 0
3 Louis 1 1 0
4 Sara 2 1 1
5 Sophie 2 1 1
6 John 4 1 2
4 Sara 2 2 0
5 Sophie 2 2 0
6 John 4 2 1
6 John 4 3 0