30👍
If you’re using a database that supports recursive common table expressions (e.g. PostgreSQL), this is precisely the use-case.
team = StaffMember.objects.raw('''
WITH RECURSIVE team(id, supervisor) AS (
SELECT id, supervisor
FROM staff_member
WHERE id = 42
UNION ALL
SELECT sm.id, sm.supervisor
FROM staff_member AS sm, team AS t
WHERE sm.id = t.supervisor
)
SELECT * FROM team
''')
References:
Raw SQL queries in Django
Recursive Common Table Expressions in PostgreSQL
3👍
I found a solution to the problem. The recursive solution takes the node, goes to it’s first child and goes deep down till bottom of the hierarchy. Then comes back up again to the second child (if exists), and then again goes down till the bottom. In short, it explores all the nodes one by one and appends all the members in an array. The solution I came up with, fetches the members layer-wise.
member = StaffMember.objects.get(id__id=user_id)
new_list = [member]
new_list = get_final_team(new_list)
def get_final_team(qs):
team = []
staffmembers = StaffMember.objects.filter(supervisor__in=qs)
team += staffmembers
if staffmembers:
interim_team_qs = get_final_team(staffmembers)
for qs in interim_team_qs:
team.append(qs)
else:
team = [qs]
return team
The number of db calls this method entails is the number of layers (of hierarchy) that are present beneath the member whose team we want to find out.
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