sql-2-afternoon

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sql-2-afternoon

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A sql afternoon project to solidify using more intermediate queries.

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# Project Summary

In this project, we'll continue to use <a href="https://postgres.devmountain.com/">postgres.devmountain.com</a> to practice joins, nested queries, updating rows, group by, distinct, and foreign keys.

Any new tables or records that you add into the database will be removed after you refresh the page.

## Practice joins

### Instructions

<summary> <code> Syntax Hint </code> </summary>

```sql
SELECT [Column names] 
FROM [table] [abbv]
JOIN [table2] [abbv2] ON abbv.prop = abbv2.prop WHERE [Conditions];

SELECT a.name, b.name FROM some_table a JOIN another_table b ON a.some_id = b.some_id;
SELECT a.name, b.name FROM some_table a JOIN another_table b ON a.some_id = b.some_id WHERE b.email = 'e@mail.com';
```

</details>

1. Get all invoices where the `unit_price` on the `invoice_line` is greater than $0.99.
2. Get the `invoice_date`, customer `first_name` and `last_name`, and `total` from all invoices.
3. Get the customer `first_name` and `last_name` and the support rep's `first_name` and `last_name` from all customers. 
    * Support reps are on the employee table.
4. Get the album `title` and the artist `name` from all albums.
5. Get all playlist_track track_ids where the playlist `name` is Music.
6. Get all track `name`s for `playlist_id` 5.
7. Get all track `name`s and the playlist `name` that they're on ( 2 joins ).
8. Get all track `name`s and album `title`s that are the genre `Alternative & Punk` ( 2 joins ).

### Solution

<summary> <code> SQL Solutions </code> </summary>

```sql
SELECT *
FROM invoice i
JOIN invoice_line il ON il.invoice_id = i.invoice_id
WHERE il.unit_price > 0.99;
```

</details>

```sql
SELECT i.invoice_date, c.first_name, c.last_name, i.total
FROM invoice i
JOIN customer c ON i.customer_id = c.customer_id;
```

</details>

```sql
SELECT c.first_name, c.last_name, e.first_name, e.last_name
FROM customer c
JOIN employee e ON c.support_rep_id = e.employee_id;
```

</details>

```sql
SELECT al.title, ar.name
FROM album al
JOIN artist ar ON al.artist_id = ar.artist_id;
```

</details>

```sql
SELECT pt.track_id
FROM playlist_track pt
JOIN playlist p ON p.playlist_id = pt.playlist_id
WHERE p.name = 'Music';
```

</details>

```sql
SELECT t.name
FROM track t
JOIN playlist_track pt ON pt.track_id = t.track_id
WHERE pt.playlist_id = 5;
```

</details>

```sql
SELECT t.name, p.name
FROM track t
JOIN playlist_track pt ON t.track_id = pt.track_id
JOIN playlist p ON pt.playlist_id = p.playlist_id;
```

</details>

```sql
SELECT t.name, a.title
FROM track t
JOIN album a ON t.album_id = a.album_id
JOIN genre g ON g.genre_id = t.genre_id
WHERE g.name = 'Alternative & Punk';
```

</details>

### Black Diamond

* Get all tracks on the playlist(s) called Music and show their name, genre name, album name, and artist name.
  * At least 5 joins.

## Practice nested queries

### Summary

Complete the instructions without using any joins. Only use nested queries to come up with the solution.

### Instructions

<summary> <code> Syntax Hint </code> </summary>

```sql
SELECT [column names] 
FROM [table] 
WHERE column_id IN ( SELECT column_id FROM [table2] WHERE [Condition] );

SELECT name, Email FROM Athlete WHERE AthleteId IN ( SELECT PersonId FROM PieEaters WHERE Flavor='Apple' );
```

</details>

1. Get all invoices where the `unit_price` on the `invoice_line` is greater than $0.99.
2. Get all playlist tracks where the playlist name is Music.
3. Get all track names for `playlist_id` 5.
4. Get all tracks where the `genre` is Comedy.
5. Get all tracks where the `album` is Fireball.
6. Get all tracks for the artist Queen ( 2 nested subqueries ).

### Solution

<summary> <code> SQL Solutions </code> </summary>

```sql
SELECT *
FROM invoice
WHERE invoice_id IN ( SELECT invoice_id FROM invoice_line WHERE unit_price > 0.99 );
```

</details>

```sql
SELECT *
FROM playlist_track
WHERE playlist_id IN ( SELECT playlist_id FROM playlist WHERE name = 'Music' );
```

</details>

```sql
SELECT name
FROM track
WHERE track_id IN ( SELECT track_id FROM playlist_track WHERE playlist_id = 5 );
```

</details>

```sql
SELECT *
FROM track
WHERE genre_id IN ( SELECT genre_id FROM genre WHERE name = 'Comedy' );
```

</details>

```sql
SELECT *
FROM track
WHERE album_id IN ( SELECT album_id FROM album WHERE title = 'Fireball' );
```

</details>

```sql
SELECT *
FROM track
WHERE album_id IN ( 
  SELECT album_id FROM album WHERE artist_id IN ( 
    SELECT artist_id FROM artist WHERE name = 'Queen'
  )
); 
```

</details>

## Practice updating Rows

### Instructions

<summary> <code> Syntax Hint </code> </summary>

```sql
UPDATE [table] 
SET [column1] = [value1], [column2] = [value2] 
WHERE [Condition];

UPDATE athletes SET sport = 'Picklball' WHERE sport = 'pockleball';
```

</details>

1. Find all customers with fax numbers and set those numbers to `null`.
2. Find all customers with no company (null) and set their company to `"Self"`.
3. Find the customer `Julia Barnett` and change her last name to `Thompson`.
4. Find the customer with this email `luisrojas@yahoo.cl` and change his support rep to `4`.
5. Find all tracks that are the genre `Metal` and have no composer. Set the composer to `"The darkness around us"`.
6. Refresh your page to remove all database changes.

### Solution

<summary> <code> SQL Solutions </code> </summary>

```sql
UPDATE customer
SET fax = null
WHERE fax IS NOT null;
```

</details>

```sql
UPDATE customer
SET company = 'Self'
WHERE company IS null;
```

</details>

```sql
UPDATE customer 
SET last_name = 'Thompson' 
WHERE first_name = 'Julia' AND last_name = 'Barnett';
```

</details>

```sql
UPDATE customer
SET support_rep_id = 4
WHERE email = 'luisrojas@yahoo.cl';
```

</details>

```sql
UPDATE track
SET composer = 'The darkness around us'
WHERE genre_id = ( SELECT genre_id FROM genre WHERE name = 'Metal' )
AND composer IS null;
```

</details>

## Group by

### Instructions

<summary> <code> Syntax Hint </code> </summary>

```sql
SELECT [column1], [column2]
FROM [table] [abbr]
GROUP BY [column];
```

</details>

1. Find a count of how many tracks there are per genre. Display the genre name with the count.
2. Find a count of how many tracks are the `"Pop"` genre and how many tracks are the `"Rock"` genre.
3. Find a list of all artists and how many albums they have.

### Solution

<summary> <code> SQL Solutions </code> </summary>

```sql
SELECT COUNT(*), g.name
FROM track t
JOIN genre g ON t.genre_id = g.genre_id
GROUP BY g.name;
```

</details>

```sql
SELECT COUNT(*), g.name
FROM track t
JOIN genre g ON g.genre_id = t.genre_id
WHERE g.name = 'Pop' OR g.name = 'Rock'
GROUP BY g.name;
```

</details>

```sql
SELECT ar.name, COUNT(*)
FROM album al
JOIN artist ar ON ar.artist_id = al.artist_id
GROUP BY ar.name;
```

</details>

## Use Distinct

<summary> <code> Syntax Hint </code> </summary>

```sql
SELECT DISTINCT [column]
FROM [table];
```

</details>

1. From the `track` table find a unique list of all `composer`s.
2. From the `invoice` table find a unique list of all `billing_postal_code`s.
3. From the `customer` table find a unique list of all `company`s.

<summary> <code> SQL Solutions </code> </summary>

```sql
SELECT DISTINCT composer
FROM track;
```

</details>

```sql
SELECT DISTINCT billing_postal_code
FROM invoice;
```

</details>

```sql
SELECT DISTINCT company
FROM customer;
```

</details>

## Delete Rows

### Summary

Always do a select before a delete to make sure you get back exactly what you want and only what you want to delete! Since we cannot delete anything from the pre-defined tables ( foreign key restraints ), use the following SQL code to create a dummy table:

<summary> <code> practice_delete TABLE </code> </summary>

```sql
CREATE TABLE practice_delete ( name TEXT, type TEXT, value INTEGER );
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'bronze', 50);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'bronze', 50);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'bronze', 50);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'silver', 100);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'silver', 100);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'gold', 150);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'gold', 150);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'gold', 150);
INSERT INTO practice_delete ( name, type, value ) VALUES ('delete', 'gold', 150);

SELECT * FROM practice_delete;
```

</details>

### Instructions

<summary> <code> Syntax Hint </code> </summary>

```sql
DELETE FROM [table] WHERE [condition]
```

</details>

1. Copy, paste, and run the SQL code from the summary.
2. Delete all `'bronze'` entries from the table.
3. Delete all `'silver'` entries from the table.
4. Delete all entries whose value is equal to `150`.

### Solution

<summary> <code> SQL Solutions </code> </summary>

```sql
DELETE 
FROM practice_delete 
WHERE type = 'bronze';
```

</details>

```sql
DELETE 
FROM practice_delete 
WHERE type = 'silver';
```

</details>

```sql
DELETE 
FROM practice_delete 
WHERE value = 150;
```

</details>

## eCommerce Simulation - No Hints

### Summary

Let's simulate an e-commerce site. We're going to need users, products, and orders.

* users need a name and an email.
* products need a name and a price
* orders need a ref to product.
* All 3 need primary keys.

### Instructions

* Create 3 tables following the criteria in the summary.
* Add some data to fill up each table.
  * At least 3 users, 3 products, 3 orders.
* Run queries against your data.
  * Get all products for the first order.
  * Get all orders.
  * Get the total cost of an order ( sum the price of all products on an order ).
* Add a foreign key reference from orders to users.
* Update the orders table to link a user to each order.
* Run queries against your data.
  * Get all orders for a user.
  * Get how many orders each user has.

### Black Diamond

* Get the total amount on all orders for each user.

### Resources

* [SQL Teaching](http://www.sqlteaching.com)
* [SQL Bolt](http://www.sqlbolt.com)

</details>

## Contributions

If you see a problem or a typo, please fork, make the necessary changes, and create a pull request so we can review your changes and merge them into the master repo and branch.

## Copyright

© DevMountain LLC, 2017. Unauthorized use and/or duplication of this material without express and written permission from DevMountain, LLC is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to DevMountain with appropriate and specific direction to the original content.

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