SQL Server CTE levé vnější spojení

Pokud není známo, kolik úrovní je v hierarchii?

Pak se taková výzva často provádí prostřednictvím rekurzivního CTE.

Příklad úryvku:

--
-- Using table variables for testing reasons
--
declare @customertest table (cid int primary key, upid int);
declare @conftest table (cid int, confname varchar(6) default 'budget', confvalue int);
--
-- Sample data
--
insert into @customertest (cid, upid) values 
(1,0), (2,1), (3,1), (4,2), (5,2), (6,3), 
(7,5), (8,5), (9,8), (10,9);
insert into @conftest (cid, confvalue) values 
(1,1000), (2,700), (3,300), (4,100), (5,200), (6,300);

-- The customer that has his own budget, or not.
declare @customerID int = 10;

;with RCTE AS 
(
  --
  -- the recursive CTE starts from here. The seed records, as one could call it.
  --
  select cup.cid as orig_cid, 0 as lvl, cup.cid, cup.upid, budget.confvalue
  from @customertest as cup
  left join @conftest budget on (budget.cid = cup.cid and budget.confname = 'budget')
  where cup.cid = @customerID -- This is where we limit on the customer

  union all

  --
  -- This is where the Recursive CTE loops till it finds nothing new
  --
  select RCTE.orig_cid, RCTE.lvl+1, cup.cid, cup.upid, budget.confvalue
  from RCTE
  join @customertest as cup on (cup.cid = RCTE.upid)
  outer apply (select b.confvalue from @conftest b where b.cid = cup.cid and b.confname = 'budget') as budget
  where RCTE.confvalue is null -- Loop till a budget is found
)
select 
 orig_cid as cid, 
 confvalue
from RCTE
where confvalue is not null;    

Výsledek:

cid confvalue
--- ---------
 10       200

Btw, rekurzivní CTE používá OUTER APPLY, protože MS SQL Server tam neumožňuje použití LEFT OUTER JOIN.

A pokud je jisté, že existuje maximální hloubka 1 úrovně pro upid s rozpočtem?
Pak stačí jednoduché spojení vlevo a splynutí by stačilo.

Například:

select cup.cid, coalesce(cBudget.confvalue, upBudget.confvalue) as confvalue
from @customertest as cup
left join @conftest cBudget on (cBudget.cid = cup.cid and cBudget.confname = 'budget')
left join @conftest upBudget on (upBudget.cid = cup.upid and upBudget.confname = 'budget')
where cup.cid = 8;