DB week_1 homework

[roza8765]

No description provided.

[zoltan-gal]

Great work Razan! I'm really impressed by what a nice solution you found for the bonus question!

[zoltan-gal]

More precisely count(distinct room_no), otherwise, you will get how many times rooms were occupied. distinct keyword removes duplicates before counting so that we will get how many different rooms were occupied.

[zoltan-gal]

Very elegant solution! One more advice, it could be made more information by adding a floor number to the query.
E.g.: select trunc(room_no/100) AS floor, um(checkout_date - checkin_date) from reservations group by room_no/100;

[zoltan-gal]

Nice work Razan! I'd recommend doing some extra checks when working with joins to make sure you will get the rows joined in the expected combination. Try to check which rows in what combination you would like to see before actually joining them by selecting those first without a join. You can also make a scratch on paper what is your expectation. Finally, always ascertaina how many rows your join should returns, you can select a smaller subset initially as an example that is managable mentally.

I'm also looking forward to seeing the 3rd exercise from this HW pack.

[zoltan-gal]

It's fine but would be nice to see product info as well.

[zoltan-gal]

Let's break it down

• Retrieve all orders, including order items for the customer 1
• Into the result, include:
  • order id,
  • order reference,
  • order date
  • total cost of the order
    that is calculated as sum of (product quantity * product unit price).

[zoltan-gal]

You're on the right track. Just need to add some extra column such as order id and the figure to calculate to end price by grouping the same order.

[zoltan-gal]

I assume order items refers to the order_items table.

[zoltan-gal]

It might be more intuitive joining the product availability through the product table and also add a join condition for connecting product on product availability table (product_availability a on (a.prod_id = p.id and a.supp_id = i.supplier_id)) to ensure you get the right combination, otherwise, join product_availability a on (i.supplier_id = a.supp_id) will join non-corresponding product rows that are still from the same suppliers.

e.g.

select p.product_name, a.unit_price, i.quantity
  from orders o join
       order_items i on (i.order_id = o.id) join
       products p on (i.product_id = p.id) join
       product_availability a on (a.prod_id = p.id and a.supp_id = i.supplier_id)
  where o.order_reference = 'ORD006';

[zoltan-gal]

It's okay but as an improvement, it is worth adding a distinct to remove duplicates if we are not interested in how many orders a customer has had.

select distinct c.name
  from customers c join
       orders o on (o.customer_id = c.id) join
       order_items i on (i.order_id = o.id) join
       suppliers s on (i.supplier_id = s.id)
  where s.country = 'China';

[zoltan-gal]

Basically, you did it here what hadn't understood here:

8. Retrieve all orders, including order items, from customer ID 1. Include order id, reference, date and total cost (calculated as quantity * unit price).

select o.order_reference, o.order_date from orders o join order_items i on (i.order_id = o.id) join product_availability a on (a.prod_id = i.product_id) where o.customer_id = 1;
////////////I don't understand what they exactly want from this question ////////////////

The only problem with the query is the same as the aforementioned.
You should and also add a join condition for connecting products on product availability table (product_availability a on (a.prod_id = p.id and a.supp_id = i.supplier_id)) to ensure you get the right combination from that table, otherwise, it will join more rows than necessary and the result will be incorrect.

select c.name, o.order_reference, o.order_date, sum(i.quantity * a.unit_price) as total
 from customers c join
      orders o on (o.customer_id = c.id) join
      order_items i on (i.order_id = o.id) join
      product_availability a on (i.supplier_id = a.supp_id and i.product_id = a.prod_id)
 group by c.name, o.order_reference, o.order_date
 order by total desc;