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README.md

@@ -35,23 +35,38 @@ 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';
+
+"customer_id","company_name","contact_name","contact_title","address","city","region","postal_code","country","phone","fax"
+"AROUT","Around the Horn","Thomas Hardy","Sales Representative","120 Hanover Sq.","London",NULL,"WA1 1DP","UK","(171) 555-7788","(171) 555-6750"
+"BSBEV","B's Beverages","Victoria Ashworth","Sales Representative","Fauntleroy Circus","London",NULL,"EC2 5NT","UK","(171) 555-1212",NULL
+"CONSH","Consolidated Holdings","Elizabeth Brown","Sales Representative","Berkeley Gardens 12  Brewery","London",NULL,"WX1 6LT","UK","(171) 555-2282","(171) 555-9199"
+"EASTC","Eastern Connection","Ann Devon","Sales Agent","35 King George","London",NULL,"WX3 6FW","UK","(171) 555-0297","(171) 555-3373"
+"NORTS","North/South","Simon Crowther","Sales Associate","South House 300 Queensbridge","London",NULL,"SW7 1RZ","UK","(171) 555-7733","(171) 555-2530"
+"SEVES","Seven Seas Imports","Hari Kumar","Sales Manager","90 Wadhurst Rd.","London",NULL,"OX15 4NB","UK","(171) 555-1717","(171) 555-5646"
 
 
 ### 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 supplier_id, company_name, city
+from suppliers
+where length(company_name) > 20
 
 ### find all customers that include the word 'MARKET' in the contact title. Should return 19 records.
 > This can be done with SELECT and a WHERE clause using the LIKE keyword
@@ -60,6 +75,8 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
 
 > Remember to convert your contact title to all upper case for case insenstive comparing so upper(contact_title)
 
+SELECT * from customers where upper(contact_title) like '%MARKET%'
+
 
 ### add a customer record for   
 * customer id is 'SHIRE'
@@ -71,24 +88,49 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
 * 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
 
+update customers
+set postal_code = '11122'
+where customer_id = 'SHIRE'
+
 
 ### list orders grouped by customer showing the number of orders per customer. _Rattlesnake Canyon Grocery_ should have 18 orders.
 > 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 count(o.order_id), c.contact_name, c.company_name 
+from orders o join customers c 
+on o.customer_id = c.customer_id
+group by c.contact_name, c.company_name
+order by count(o.order_id)
+
 
 ### 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 count(o.order_id), c.contact_name, c.company_name
+from orders o join customers c 
+on o.customer_id = c.customer_id
+group by c.contact_name, c.company_name
+order by count(o.order_id) 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 count(o.order_id), c.city
+from customers c join orders o
+on o.customer_id = c.customer_id
+group by c.city
+order by c.city
+
 
 ## Data Normalization
 
@@ -102,6 +144,19 @@ Take the following data and normalize it into a 3NF database.
 | Bob         | Joe      | Horse    |            |            |            |            | No          | No           |
 | Sam         | Ginger   | Dog      | Miss Kitty | Cat        | Bubble     | Fish       | Yes         | No           |
 
+
+Person Name|	Fenced Yard|	City Dweller|	Person Id
+Jane	No	Yes	1
+Bob	No	No	2
+Sam	Yes	No	3
+Person Id|	Pet Id|	Pet Name|	Pet Type|
+1	1	Ellie	Dog
+1	2	Tiger	Cat
+1	3	Toby	Turtle
+2	4	Joe	Horse
+3	5	Ginger	Dog
+3	6	Miss Kitty	Cat
+3	7	Bubble	Fish
 ---
 ## Stretch Goals