1. Explain the importance of grouping and sorting data stored in a database
2. Group and sort data with the GROUP BY() and ORDER BY() keywords
3. Craft advanced queries using aggregator functions along with sorting keywords and other conditional clauses

SQL isn't picky about how it returns data to you, based on your queries. It will simply return the relevant table rows in the order in which they exist in the table. This is often insufficient for the purposes of data analysis and organization.

How common is it to order a list of items alphabetically? Or numerically from least to greatest?

We can tell our SQL queries and aggregate functions to group and sort our data using a number of clauses:

• ORDER BY()
• GROUP BY()
• HAVING
• ASC/DESC
• LIMIT

Let's take a closer look at how we use these keywords to narrow our search criteria as well as to order and group it.

Some cats are very famous, and accordingly very wealthy. Our Pet's Database will have a Cats table in which each cat has a name, age, breed and net worth. Our database will also have an Owners table and cats_owners join table so that a cat can have many owners and an owner can have many cats.

Creating the Database:

Create the database in your terminal with the following:

sqlite3 pets_database.db

Creating the tables:

In the sqlite3> prompt in your terminal:

Cats table:

CREATE TABLE cats (
id INTEGER PRIMARY KEY,
name TEXT,
age INTEGER,
breed TEXT, 
net_worth INTEGER
);

Owners Table:

CREATE TABLE owners (id INTEGER PRIMARY KEY, name TEXT);

Cats Owners Table:

CREATE TABLE cats_owners (
cat_id INTEGER,
owner_id INTEGER
);

Inserting the values:

Cats:

INSERT INTO cats (id, name, age, breed, net_worth) VALUES (a, "Maru", 3, "Scottish Fold", 1000000);
INSERT INTO cats (id, name, age, breed, net_worth) VALUES (2, "Hana", 1, "Tabby", 21800);
INSERT INTO cats (id, name, age, breed, net_worth) VALUES (3, "Grumpy Cat", 4, "Persian", 181600);
INSERT INTO cats (id, name, age, breed, net_worth) VALUES (4, "Lil' Bub", 2, "Tortoiseshell", 200000);

Owners:

INSERT INTO owners (name) VALUES ("mugumogu");
INSERT INTO owners (name) VALUES ("Sophie");
INSERT INTO owners (name) VALUES ("Penny");

Cats Owners:

INSERT INTO cats_owners (cat_id, owner_id) VALUES (3, 2);
INSERT INTO cats_owners (cat_id, owner_id) VALUES (3, 3);
INSERT INTO cats_owners (cat_id, owner_id) VALUES (1, 2);

SELECT column_name, column_name
FROM table_name
ORDER BY column_name ASC|DESC, column_name ASC|DESC;

ORDER BY() will automatically sort the returned values in ascending order. Use the DESC keyword, as above, to sort in descending order.

Imagine you're working for an important investment firm in Manhattan. The investors are interested in investing in a lucrative and popular cat. They need your help to decide which cat that will be. They want a list of famous and wealthy cats. We can do that with a basic SELECT statement:

SELECT * FROM cats WHERE net_worth > 0;

This will return:

name             age         breed          net_worth 
---------------  ----------  -------------  ----------
Maru             3           Scottish Fold  1000000   
Hana             1           Tabby          21800     
Grumpy Cat       4           Persian        181600    
Lil' Bub         2           Tortoiseshell  2000000   

Our investors are busy people though. They don't have time to manually sort through this list of cats for the best candidate. They want you to return the list to them with the cats sorted by net worth, from greatest to least.

We can do so with the following lines:

SELECT * FROM cats ORDER BY(net_worth) DESC;

This will return:

name             age         breed          net_worth 
---------------  ----------  -------------  ----------
Lil' Bub         2           Tortoiseshell  2000000   
Maru             3           Scottish Fold  1000000   
Grumpy Cat       4           Persian        181600    
Hana             1           Tabby          21800     
Nona             4           Tortoiseshell         

Turns out our investors are very impatient. They don't want to review the list themselves, they just want you to return to them the wealthiest cat. We can accomplish this by using the LIMIT keyword with the above query:

SELECT * FROM cats ORDER BY(net_worth) DESC LIMIT 1;
name             age         breed          net_worth 
---------------  ----------  -------------  ----------
Lil' Bub         2           Tortoiseshell  2000000   

The LIMIT keyword specifies how many of records that result from the query you'd like to actually return.

The GROUP BY() keyword is very similar to ORDER BY(). The only difference is that ORDER BY() sorts the resulting data set of basic queries while GROUP BY() sorts the result sets of aggregate functions.

SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name;

Let's calculate the sum of the net worth of all of the cats that belong to our second owner:

SELECT SUM(Cats.net_worth) 
FROM Owners 
INNER JOIN cats_owners 
ON Owners.id = cats_owners.owner_id 
JOIN Cats ON cats_owners.cat_id = Cats.id 
WHERE cats_owners.owner_id = 2;

This should return:

SUM(Cats.net_worth) 
--------------------
1181600 

In the above query, we use the SUM(Cats.net_worth) aggregator. SUM looks at the all of the values in the net_worth column of the Cats table (or whatever column you specify in parentheses) and takes the sum of the those values.