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44 changes: 38 additions & 6 deletions README.md
Original file line number Diff line number Diff line change
Expand Up @@ -35,22 +35,36 @@ Answer the following data queries. Keep track of the SQL you write by pasting it

### find all customers that live in London. Returns 6 records.
> This can be done with SELECT and WHERE clauses

SELECT *
FROM customers
WHERE city = 'London'

### find all customers with postal code 1010. Returns 3 customers.
> This can be done with SELECT and WHERE clauses

SELECT *
FROM customers
WHERE postal_code = '1010'

### find the phone number for the supplier with the id 11. Should be (010) 9984510.
> This can be done with SELECT and WHERE clauses
SELECT phone
FROM suppliers
WHERE supplier_id = '11'


### list orders descending by the order date. The order with date 1998-05-06 should be at the top.
> This can be done with SELECT, WHERE, and ORDER BY clauses

SELECT *
From orders
ORDER BY order_date DESC


### find all suppliers who have names longer than 20 characters. You can use `length(company_name)` to get the length of the name. Returns 11 records.
> This can be done with SELECT and WHERE clauses
SELECT *
FROM suppliers
WHERE length(company_name) > '20'


### find all customers that include the word 'MARKET' in the contact title. Should return 19 records.
Expand All @@ -59,6 +73,9 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
> Don't forget the wildcard '%' symbols at the beginning and end of your substring to denote it can appear anywhere in the string in question

> Remember to convert your contact title to all upper case for case insenstive comparing so upper(contact_title)
SELECT *
FROM customers
WHERE contact_title LIKE '%Market%'


### add a customer record for
Expand All @@ -70,7 +87,8 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
* the postal code is '111'
* the country is 'Middle Earth'
> This can be done with the INSERT INTO clause

INSERT INTO customers(customer_id, company_name, contact_name, address, city, postal_code, country)
VALUES ('SHIRE', 'The Shire', 'Bilbo Baggins', '1 Hobbit-Hole', 'Bag End', 111, 'Middle Earth' )

### update _Bilbo Baggins_ record so that the postal code changes to _"11122"_.
> This can be done with UPDATE and WHERE clauses
Expand All @@ -80,15 +98,29 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
> This can be done with SELECT, COUNT, JOIN and GROUP BY clauses. Your count should focus on a field in the Orders table, not the Customer table

> There is more information about the COUNT clause on [W3 Schools](https://www.w3schools.com/sql/sql_count_avg_sum.asp)

SELECT c.company_name, COUNT(*) AS numOfOrders
FROM orders o
LEFT JOIN customers c
ON o.customer_id = c.customer_id
GROUP BY c.company_name

### list customers names and the number of orders per customer. Sort the list by number of orders in descending order. _Save-a-lot Markets should be at the top with 31 orders followed by _Ernst Handle_ with 30 orders. Last should be _Centro comercial Moctezuma_ with 1 order.
> This can be done by adding an ORDER BY clause to the previous answer

SELECT c.company_name, COUNT(*) AS numOfOrders
FROM orders o
LEFT JOIN customers c
ON o.customer_id = c.customer_id
GROUP BY c.company_name
ORDER BY numOfOrders DESC

### list orders grouped by customer's city showing number of orders per city. Returns 69 Records with _Aachen_ showing 6 orders and _Albuquerque_ showing 18 orders.
> This is very similar to the previous two queries, however, it focuses on the City rather than the CustomerName

SELECT c.city, COUNT(*) AS numOfOrders
FROM orders o
LEFT JOIN customers c
ON o.customer_id = c.customer_id
GROUP BY c.city
ORDER BY numOfOrders DESC

## Data Normalization

Expand Down