Optymalizacja SQL Case statement

Question

Wierzę, że mogę zoptymalizować to wyrażenie sql, używając instrukcji case dla Left Outer Joins.

Ale miałem trudny czas ustawiania skrzynek, jeden do podsumowania typów kodów AB, CD i inny dla wszystkich pozostałych.

Doceniam każdą pomoc lub wskazówki, które możesz mi w tej sprawie udzielić.

update billing set payments = isnull(bd1.amount, payments) 
, payments = case 
     when payments is null then 0 
     else payments 
    end 
, charges = case 
     when bd2.amount is not null then charges 
     when charges is null then 0 
     else charges 
     end 
, balance = round(charges + isnull(bd1.amount, bi.payments), 2) 
from billing bi 

left outer join (select inv, round(sum(bd1.bal), 2) amount 
       from "bill" bd1 
       where code_type = 'AB' 
       or code_type = 'CD' 
       group by inv) bd1 
       on bd1.inv = bi.inv 
left outer join (select invoice, round(sum(bd2.bal), 2) amount 
       from "bill" bd2 
       where code_type <> 'AB' 
       and code_type <> 'CD' 
       group by inv) bd2 
       on bd2.inv = bi.inv; 

Answer 1

Można to uprościć, aby użyć pojedynczego zapytania zamiast dwóch. Nadal potrzebujesz tego, ponieważ GROUP BY w UPDATE nie działa.

UPDATE bi 
SET payments = bd.payments, 
     charges= bd.charges, 
     balance = bd.balance 
FROM billing bi 
     LEFT JOIN (SELECT bd.inv, 
         payments = Round(Sum(CASE 
               WHEN code_type IN ('AB' , 'CD') THEN 
               bd.bal 
               ELSE 0 
               END), 2), 
         charges = Round(Sum(CASE 
               WHEN code_type NOT IN ('AB' , 'CD') THEN 
               bd.bal 
               ELSE 0 
              END), 2), 
         balance = Round(Sum(bd.bal), 2) 
        FROM bill bd 
        GROUP BY bd.inv) bd 
     ON bd.inv = bi.inv 

Answer 2

Może coś takiego:

update billing set payments = isnull(bd1.amount, payments) 
, payments = isnull(payments, 0) 
, charges = isnull(bd2.amount, isnull(charges, 0)) 
, balance = round(charges + isnull(bd1.amount, bi.payments), 2) 
from billing bi 

left outer join (select inv, round(sum(bd1.bal), 2) amount 
       from "bill" bd1 
       where code_type in ('AB', 'CD') 
       group by inv) bd1 
       on bd1.inv = bi.inv 
left outer join (select invoice, round(sum(bd2.bal), 2) amount 
       from "bill" bd2 
       where code_type not in ('AB', 'CD') 
       group by inv) bd2 
       on bd2.inv = bi.inv; 

Dwie lewe przyłącza nie są problemem!