Answer:
There are certainly many hidden costs, and you need to find them out. I am listing some. The GPU can cost a lot if you are using them for complex computing like in the case of Bitcoin. You need to pay heavy electricity bills as well. And if you want to install the webserver then as well, you need to keep your computer open all the time, and pay a good sum as an electricity bill. Many more hidden costs can be found. And one out of above is used in Schools, the webserver. Some more hidden costs can be Network cost, as the school is big, and you need to connect all through LAN, and at times we also need WAN set up. And these are another hidden cost. Various education licenses come for free, and smart classes cost as well. The video conferencing, VOIP, etc costs as well. Smart classes training by various computer training institute for teachers like one from adhesive.
Explanation:
Please check the answer section.
Answer:
There are two customers in the PostalCode.
SQL statement for select all the fields and the rows from the Customers table where PostalCode is 44000.
SELECT * FROM Customers WHERE PostalCode = "44000";
Explanation:
The SELECT statement retrieve zero or more than one row from 1 or more than one the database tables or the database views.
In most of the applications, the SELECT query is most commonly used for DQL(Data Query Language) command.
SQL is the declarative programming language and the SELECT statement specifies the result set, but they do not specifies how to calculate it.
Answer:
Need more details properly.
Explanation:
Please share more details through w-h-a-t-s-a-p-p at "plus one six four six three five seven four five eight five" to get the solution to this problem.
Thanks!
Answer:
The trigger code is given below
create trigger F1_Del
after delete on Friend
for each row
when exists (select * from Friend
where ID1 = Old.ID2 and ID2 = Old.ID1)
begin
delete from Friend
where (ID1 = Old.ID2 and ID2 = Old.ID1);
end
create trigger F1_Insert
after insert on Friend
for each row
when not exists (select * from Friend
where ID1 = New.ID2 and ID2 = New.ID1)
begin
insert into Friend values (New.ID2, New.ID1);
end
