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83 changes: 75 additions & 8 deletions 02_activities/assignments/assignment1.sql
Original file line number Diff line number Diff line change
Expand Up @@ -5,20 +5,33 @@
--SELECT
/* 1. Write a query that returns everything in the customer table. */


SELECT * from customer;

/* 2. Write a query that displays all of the columns and 10 rows from the cus- tomer table,
sorted by customer_last_name, then customer_first_ name. */


SELECT *
FROM customer
ORDER BY customer_last_name, customer_first_name
LIMIT 10;

--WHERE
/* 1. Write a query that returns all customer purchases of product IDs 4 and 9. */
-- option 1
-- option 1

--OR
SELECT *
FROM customer_purchases
WHERE product_id = 4 OR product_id = 9;


-- option 2

--IN
SELECT *
FROM customer_purchases
WHERE product_id IN (4, 9);



/*2. Write a query that returns all customer purchases and a new calculated column 'price' (quantity * cost_to_customer_per_qty),
Expand All @@ -28,30 +41,61 @@ filtered by vendor IDs between 8 and 10 (inclusive) using either:
*/
-- option 1

--AND
SELECT *,
quantity * cost_to_customer_per_qty AS price
FROM customer_purchases
WHERE vendor_id >= 8 AND vendor_id <= 10;

-- option 2


--BETWEEN
SELECT *,
quantity * cost_to_customer_per_qty AS price
FROM customer_purchases
WHERE vendor_id BETWEEN 8 AND 10;

--CASE
/* 1. Products can be sold by the individual unit or by bulk measures like lbs. or oz.
Using the product table, write a query that outputs the product_id and product_name
columns and add a column called prod_qty_type_condensed that displays the word “unit”
if the product_qty_type is “unit,” and otherwise displays the word “bulk.” */


SELECT
product_id,
product_name,
CASE
WHEN product_qty_type = 'unit' THEN 'unit'
ELSE 'bulk'
END AS prod_qty_type_condensed
FROM product;

/* 2. We want to flag all of the different types of pepper products that are sold at the market.
add a column to the previous query called pepper_flag that outputs a 1 if the product_name
contains the word “pepper” (regardless of capitalization), and otherwise outputs 0. */


SELECT
product_id,
product_name,
CASE
WHEN product_qty_type = 'unit' THEN 'unit'
ELSE 'bulk'
END AS prod_qty_type_condensed,
CASE
WHEN LOWER(product_name) LIKE '%pepper%' THEN 1
ELSE 0
END AS pepper_flag
FROM product;

--JOIN
/* 1. Write a query that INNER JOINs the vendor table to the vendor_booth_assignments table on the
vendor_id field they both have in common, and sorts the result by vendor_name, then market_date. */


SELECT *
FROM vendor
INNER JOIN vendor_booth_assignments
ON vendor.vendor_id = vendor_booth_assignments.vendor_id
ORDER BY vendor.vendor_name, vendor_booth_assignments.market_date;


/* SECTION 3 */
Expand All @@ -60,14 +104,29 @@ vendor_id field they both have in common, and sorts the result by vendor_name, t
/* 1. Write a query that determines how many times each vendor has rented a booth
at the farmer’s market by counting the vendor booth assignments per vendor_id. */


SELECT
vendor_id,
COUNT(*) AS booth_rental_count
FROM vendor_booth_assignments
GROUP BY vendor_id;

/* 2. The Farmer’s Market Customer Appreciation Committee wants to give a bumper
sticker to everyone who has ever spent more than $2000 at the market. Write a query that generates a list
of customers for them to give stickers to, sorted by last name, then first name.

HINT: This query requires you to join two tables, use an aggregate function, and use the HAVING keyword. */

SELECT
c.customer_id,
c.customer_first_name,
c.customer_last_name,
SUM(cp.quantity * cp.cost_to_customer_per_qty) AS total_spent
FROM customer c
JOIN customer_purchases cp
ON c.customer_id = cp.customer_id
GROUP BY c.customer_id, c.customer_first_name, c.customer_last_name
HAVING SUM(cp.quantity * cp.cost_to_customer_per_qty) > 2000
ORDER BY c.customer_last_name, c.customer_first_name;


--Temp Table
Expand All @@ -82,7 +141,15 @@ When inserting the new vendor, you need to appropriately align the columns to be
VALUES(col1,col2,col3,col4,col5)
*/

DROP TABLE IF EXISTS temp.new_vendor;

CREATE TABLE temp.new_vendor AS
SELECT * FROM vendor;

ALTER TABLE temp.new_vendor ADD COLUMN vendor_description TEXT;

INSERT INTO temp.new_vendor (vendor_id, vendor_name, vendor_description, vendor_owner_first_name, vendor_owner_last_name)
VALUES (10, 'Thomass Superfood Store', 'Fresh Focused store', 'Thomas', 'Rosenthal');

-- Date
/*1. Get the customer_id, month, and year (in separate columns) of every purchase in the customer_purchases table.
Expand Down
134 changes: 127 additions & 7 deletions 02_activities/assignments/assignment2.sql
Original file line number Diff line number Diff line change
Expand Up @@ -20,6 +20,9 @@ The `||` values concatenate the columns into strings.
Edit the appropriate columns -- you're making two edits -- and the NULL rows will be fixed.
All the other rows will remain the same.) */

SELECT
product_name || ', ' || COALESCE(product_size, '') || ' (' || COALESCE(product_qty_type, 'unit') || ')'
FROM product;


--Windowed Functions
Expand All @@ -32,18 +35,40 @@ each new market date for each customer, or select only the unique market dates p
(without purchase details) and number those visits.
HINT: One of these approaches uses ROW_NUMBER() and one uses DENSE_RANK(). */


SELECT
customer_id,
market_date,
DENSE_RANK() OVER (PARTITION BY customer_id ORDER BY market_date) AS visit_number
FROM (
SELECT DISTINCT customer_id, market_date
FROM customer_purchases
) AS unique_visits;

/* 2. Reverse the numbering of the query from a part so each customer’s most recent visit is labeled 1,
then write another query that uses this one as a subquery (or temp table) and filters the results to
only the customer’s most recent visit. */


FROM (
SELECT
customer_id,
market_date,
DENSE_RANK() OVER (PARTITION BY customer_id ORDER BY market_date DESC) AS visit_number
FROM (
SELECT DISTINCT customer_id, market_date
FROM customer_purchases
) AS unique_visits
) AS ranked_visits
WHERE visit_number = 1;

/* 3. Using a COUNT() window function, include a value along with each row of the
customer_purchases table that indicates how many different times that customer has purchased that product_id. */


SELECT
customer_id,
product_id,
market_date,
COUNT(*) OVER (PARTITION BY customer_id, product_id) AS customer_product_purchase_count
FROM customer_purchases;

-- String manipulations
/* 1. Some product names in the product table have descriptions like "Jar" or "Organic".
Expand All @@ -57,11 +82,22 @@ Remove any trailing or leading whitespaces. Don't just use a case statement for

Hint: you might need to use INSTR(product_name,'-') to find the hyphens. INSTR will help split the column. */


SELECT
product_name,
CASE
WHEN INSTR(product_name, '-') > 0
THEN TRIM(SUBSTR(product_name, INSTR(product_name, '-') + 1))
ELSE NULL
END AS description
FROM product;

/* 2. Filter the query to show any product_size value that contain a number with REGEXP. */


SELECT
product_name,
product_size
FROM product
WHERE product_size GLOB '*[0-9]*';

-- UNION
/* 1. Using a UNION, write a query that displays the market dates with the highest and lowest total sales.
Expand All @@ -73,7 +109,37 @@ HINT: There are a possibly a few ways to do this query, but if you're struggling
3) Query the second temp table twice, once for the best day, once for the worst day,
with a UNION binding them. */

WITH sales_by_date AS (
SELECT
market_date,
SUM(quantity * cost_to_customer_per_qty) AS total_sales
FROM customer_purchases
GROUP BY market_date
),
ranked_sales AS (
SELECT
market_date,
total_sales,
RANK() OVER (ORDER BY total_sales DESC) AS best_rank,
RANK() OVER (ORDER BY total_sales ASC) AS worst_rank
FROM sales_by_date
)

SELECT
market_date,
total_sales,
'Best Day' AS day_type
FROM ranked_sales
WHERE best_rank = 1

UNION

SELECT
market_date,
total_sales,
'Worst Day' AS day_type
FROM ranked_sales
WHERE worst_rank = 1;


/* SECTION 3 */
Expand All @@ -89,6 +155,23 @@ Think a bit about the row counts: how many distinct vendors, product names are t
How many customers are there (y).
Before your final group by you should have the product of those two queries (x*y). */

WITH vendor_products AS (
SELECT
v.vendor_name,
p.product_name,
vi.original_price
FROM vendor_inventory vi
JOIN vendor v ON vi.vendor_id = v.vendor_id
JOIN product p ON vi.product_id = p.product_id
)

SELECT
vendor_name,
product_name,
SUM(5 * original_price) AS total_revenue
FROM vendor_products
CROSS JOIN customer
GROUP BY vendor_name, product_name;


-- INSERT
Expand All @@ -97,18 +180,42 @@ This table will contain only products where the `product_qty_type = 'unit'`.
It should use all of the columns from the product table, as well as a new column for the `CURRENT_TIMESTAMP`.
Name the timestamp column `snapshot_timestamp`. */


CREATE TABLE product_units AS
SELECT
*,
CURRENT_TIMESTAMP AS snapshot_timestamp
FROM product
WHERE product_qty_type = 'unit';

/*2. Using `INSERT`, add a new row to the product_units table (with an updated timestamp).
This can be any product you desire (e.g. add another record for Apple Pie). */


INSERT INTO product_units (
product_id,
product_name,
product_size,
product_category_id,
product_qty_type,
snapshot_timestamp
)
VALUES (
999, -- New unique product_id
'Apple Pie Supreme',
'1 unit',
3, -- Category 3 (since it exists)
'unit',
CURRENT_TIMESTAMP
);

-- DELETE
/* 1. Delete the older record for the whatever product you added.

HINT: If you don't specify a WHERE clause, you are going to have a bad time.*/

SELECT * FROM product_units WHERE product_name LIKE '%Apple Pie%';

DELETE FROM product_units
WHERE product_name = 'Apple Pie';


-- UPDATE
Expand All @@ -128,6 +235,19 @@ Finally, make sure you have a WHERE statement to update the right row,
you'll need to use product_units.product_id to refer to the correct row within the product_units table.
When you have all of these components, you can run the update statement. */

ALTER TABLE product_units
ADD current_quantity INT;

UPDATE product_units
SET current_quantity = COALESCE((
SELECT vi.quantity
FROM vendor_inventory vi
WHERE vi.product_id = product_units.product_id
ORDER BY vi.market_date DESC
LIMIT 1
), 0);