abrahamhall84/sql_class_exercise

In class sql practice

In class exercise - Apartment lab

• Create a database called apartmentlab
• Using this database, create two tables, one for owners and one for properties
• Keep this relationship in mind when designing your schema:
  • One owner can have many properties

###Tables

• The owners table should consist of:
  • owner_id (this should be the primary key as well as a unique number that increments automatically)
  • name
  • age
• The properties table should consist of:
  • property_id (this should be the primary key as well as a unique number that increments automatically)
  • name
  • number of units
  • owner_id (this should have the constraint NOT NULL)
    • There should be also be a foreign key that references the owners table

###Questions Write down the following sql statements that are required to solve the following tasks.

1. Show all the tables. \dt
2. Show all the users. 
3. Show all the data in the owners table. <!-- SELECT * FROM owners; -->
4. Show the names of all owners. <!-- SELECT names FROM owners; -->
5. Show the ages of all of the owners in ascending order. <!-- SELECT age FROM owners ORDER BY age ASC; -->
6. Show the name of an owner whose name is Mary. <!-- SELECT name FROM owners WHERE name='Mary'; -->
7. Show the age of all owners who are older than 30. ?????????<!-- SELECT age FROM owners WHERE age < 30; -->???????????????
8. Show the name of all owners whose name starts with an E. <!-- SELECT name FROM owners WHERE name='E%'; -->
9. Add an owner named John who is 33 years old to the owners table. <!-- INSERT INTO owners (name, age) VALUES ('John', 33); -->
10. Add an owner named Jane who is 43 years old to the owners table. <!-- INSERT INTO owners (name, age) VALUES ('Jane', 43); -->
11. Change Jane's age to 30. <!-- UPDATE owners SET age='30' WHERE name='Jane' AND age='43'; -->
12. Change Jane's name to Janet. <!-- UPDATE owners SET name='Janet' WHERE name='Jane'; -->
13. Add a property named Archstone that has 20 units. <!-- INSERT INTO properties (name, units) VALUES ('Archstone', 20, 1); -->
14. Delete the owner named Janet. <!-- DELETE FROM owners WHERE names='Janet'; -->
15. Show all of the properties in alphabetical order that are not named Archstone and do not have an id of 3 or 5. ??????????????<!-- SELECT * FROM properties ORDER BY name ASC WHERE name<> ('Archstone') AND WHERE id<>3 OR 5; --> ????????????????????????
16. Count the total number of rows in the properties table. <!-- SELECT count(*) FROM properties; -->
17. Show the highest age of all owners. <!-- SELECT age FROM owners ORDER BY age DESC LIMIT 1; -->
18. Show the names of the first three owners in your owners <table class=""></table> <!-- SELECT * FROM owners LIMIT 3; -->
19. Create a foreign key that references the owner_id in the owners table and forces the constraint ON DELETE NO ACTION. <!-- ALTER TABLE properties ADD CONSTRAINT owners_fk FOREIGN KEY (owners_id) REFERENCES owners (id) ON DELET NO ACTION; -->
20. Show all of the information from the owners table and the properties table in one joined table.
<!-- SELECT * FROM owners JOIN properties on owners.id = properties.owners_id;   -->

Bonus (this might require you to look up documentation online)

1. In the properties table change the name of the column "name" to "property_name". 
2. Count the total number of properties where the owner_id is between 1 and 3.
3. Delete the owners table - what happens? why?