/*   a. Find distinct track names that start with “Z”.Sort the output alphabetically.*/

/*select distinct track.Name
from Track where Name like 'Z%' order by Name;*/


/*   b. Find the first names of the employees who are older than their supervisor.
Hint: ReportsTo attribute in Employee table stores the EmployeeId of the supervisor. Sort the output alphabetically*/

/*select FirstName from Employee as E1
where BirthDate >
(select BirthDate from Employee where EmployeeId = E1.ReportsTo)
order by FirstName;*/


/*   c. Find the name of the highest priced track. If more than one track has the highest
price, return the names of all such tracks. Sort the output alphabetically based on the track name*/

/*select Track.name from Track
where UnitPrice >=
(select max(e.UnitPrice) from Track as e)
order by Name;*/


/*   d. Find a list containing the total amount spend by a customer. Include the customer’s id and
the last names along with the total amount. For customers who did not make any purchase, make sure to include
 them as well (the total should be 0.00 for those customers).BUG HERE DOESNT WORK HOW SUPPOSED TO PROBS*/

 /*select CustomerId, LastName, Total as Total_Spent
 from Customer Natural right outer join Invoice;*/


 /*   e. Find the title of the highest priced album CHECK IF WORKS*/

 /*select Title from Album natural join Track
 group by AlbumId having sum(UnitPrice) >= all
 (select sum(UnitPrice) from Track  group by AlbumId);*/


 /*   f. Finda distinct list containing the titles of albums that are never sold. Consider
 an album never sold if none of its tracks are sold. Sort the output alphabetically.*/

 /*select distinct Title
 from Album natural join Track natural join invoiceline
 group by AlbumId
 having sum(Quantity) = 0
 order by Title;*/


 /*   g. Create a view that returns customers’ first and last names along with
corresponding sums of all their invoice totals. Name the view as “CustomerInvoices.” */

/*create view CustomerInvoices as
select FirstName, LastName, sum(Total) as InvoiceTotal
from Customer natural join invoice
group by CustomerId;

select* from CustomerInvoices;*/