Hi,
Nice question.... I made a nice query that'll get you the result. I will not give
you the code because: 1. it's your exam, 2. I think It has some "rough edges" and
I don't want you to get the blame of my "edges" :-)
I'll give you a narrative version:
1. I'm counting the number of distinct makes...
2. I'm counting the number of distinct makes every customer has ever used (join,
join, join!) and compare that number to the result of #1
This of course happens in subqueries. I do not use HAVINGs though...
And on the design of the exam-database; lousy.
Succes,
A. Prins.
"dejauser2001@yahoo.co.uk" schreef:
> Hi everyone,
>
> Its coming close to my January examinations, and
> while revising I got stuck in the follwing question.
> I literally spent hours trying to solve it but I
> just can't :(
>
> Consider the folowing relations; (where the * means
> primary key)
>
> The "Car Rental" database
> -------------------------
>
> customer(*cust_num*,cust_name)
> hire(*car_reg*,*cust_num*,*hire_date*)
> shop(*outlet_code*,address)
> car_model(*model*,make,num_seats,max_speed)
> car(*car reg*,model,year, outlet_code)
>
> And the question is;
> Using SQL (using SELECT, FROM, WHERE, GROUP BY, COUNT(*), etc..) write a query
> to list the names of customers who have borrowed cars made by every maker.
>
> Thats it! Hard or what?
>
> Any help greatly apreciated