SQL Server trigger active card - sql

Question

I have the following tables:
Cards (IDCard, IDPerson, Balance, Active)
Persons (IDPerson)
Every time a new card is inserted, I need to check if that card already belongs to a person. If so, I need to change the active to false and the new card must receive the balance of the previous card and the state must be true.
select count(c.IDPerson)
from cards c, inserted i
where c.IDPerson = i.IDPerson;
if ##ROWCOUNT > 0
begin
print'this card already exists'
select c.IDPerson, c.Balance, c.Active
INTO #tmp
FROM Cards c, inserted i
where c.IDCard = i.IDCard
update Cards
set Active = 0
from #tmp
where #tmp.IDPerson = Cards.IDPerson;
end
When I insert a new record into cards table, the old card does turn false (0), however the new card also stays with active equal to false.
This in an after insert on cards table.
Can someone tell me how to fix the problem?

Answer 1

Although your trigger is running within the transaction, I think it is still capable of updating itself.
When you update where the INSERTED.IDPERSON = CARD.IDPERSON - this includes the record you have inserted; thus all records are updated to Active = false.
Rather, you should do
update Cards
set Active = 0
from #tmp
where #tmp.IDPerson = Cards.IDPerson
AND Cards.IDCARD != #tmp.IDCARD --This is the new condition
I've created a copy of what I believe you are asking in this DBFiddle Link
User marc_s is also correct in saying you should use JOIN syntax here. Your entire operation could be completed in one much cleaner statement
UPDATE c
SET c.Active = 0
FROM Cards c
INNER JOIN Inserted i
ON i.IDPERSON = c.IDPERSON
AND i.IDCARD != c.IDCARD

Answer 2

The if is not appropriate. Remember, that the trigger could be handling multiple rows at the same time.
An explicit join does the filtering you want. This is a little complicated, but you can do it with two joins:
update c2
set Active = (case when c2.IDCard = i.IDCard then 1 else 0 end)
from Cards c
inserted i
on c.IDCard = i.IDCard join
Cards c2
on c.IDPerson = c2.IDPerson;