BEARBEITEN
Arbeitsbeispiel zu Gordons Notiz
Abfrage berechnet den Knotenpfad, da Sie eine feste maximale Baumtiefe haben, und sortieren Sie danach.
MySQL 5.5.30-Schema-Setup :
create table mytable(id int, parent_id int, name varchar(100));
insert mytable(id, parent_id, name)
values (1, null, 'Home'),
(2, null, 'Services'),
(3, 2, 'Baking'),
(4, 3, 'Cakes'),
(5, 3, 'Bread'),
(6, 5, 'Flat Bread'),
(7, 1, 'Something');
Abfrage 1 :
select t0.*,
concat(
case coalesce(t4.Parent_ID, 0)
when 0 then ''
else concat(cast(t4.Parent_ID as char), '\\')
end,
case coalesce(t3.Parent_ID, 0)
when 0 then ''
else concat(cast(t3.Parent_ID as char), '\\')
end,
case coalesce(t2.Parent_ID, 0)
when 0 then ''
else concat(cast(t2.Parent_ID as char), '\\')
end,
case coalesce(t1.Parent_ID, 0)
when 0 then ''
else concat(cast(t1.Parent_ID as char), '\\')
end,
case coalesce(t0.Parent_ID, 0)
when 0 then ''
else concat(cast(t0.Parent_ID as char), '\\')
end,
cast(t0.id as char)
) as path
from mytable t0
left join mytable t1 on t0.Parent_ID = t1.Id
left join mytable t2 on t1.Parent_ID = t2.Id
left join mytable t3 on t2.Parent_ID = t3.Id
left join mytable t4 on t3.Parent_ID = t4.Id
order by
concat(
case coalesce(t4.Parent_ID, 0)
when 0 then ''
else concat(cast(t4.Parent_ID as char), '\\')
end,
case coalesce(t3.Parent_ID, 0)
when 0 then ''
else concat(cast(t3.Parent_ID as char), '\\')
end,
case coalesce(t2.Parent_ID, 0)
when 0 then ''
else concat(cast(t2.Parent_ID as char), '\\')
end,
case coalesce(t1.Parent_ID, 0)
when 0 then ''
else concat(cast(t1.Parent_ID as char), '\\')
end,
case coalesce(t0.Parent_ID, 0)
when 0 then ''
else concat(cast(t0.Parent_ID as char), '\\')
end,
cast(t0.id as char)
)
| ID | PARENT_ID | NAME | PATH |
-----------------------------------------
| 1 | (null) | Home | 1 |
| 7 | 1 | Something | 1\7 |
| 2 | (null) | Services | 2 |
| 3 | 2 | Baking | 2\3 |
| 4 | 3 | Cakes | 2\3\4 |
| 5 | 3 | Bread | 2\3\5 |
| 6 | 5 | Flat Bread | 2\3\5\6 |
