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Assignment2 #2

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ad1ec40
I updated the assignment 1 section 1, 2 and 3
anrojas8 Nov 18, 2025
63d173c
Added assignment 2 updates
anrojas8 Nov 25, 2025
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Binary file added 02_activities/assignments/Cohort_8/Assigment1Section1AndresRojas.pdf
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76 changes: 65 additions & 11 deletions 02_activities/assignments/Cohort_8/assignment1.sql
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Expand Up @@ -4,17 +4,23 @@

--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. */

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 @@ -23,38 +29,61 @@ filtered by customer IDs between 8 and 10 (inclusive) using either:
2. one condition using BETWEEN
*/
-- option 1

SELECT *, (quantity*cost_to_customer_per_qty) AS price
FROM customer_purchases
WHERE product_id >= 8 AND product_id <= 10

-- option 2

SELECT *, (quantity*cost_to_customer_per_qty) AS price
FROM customer_purchases
WHERE product_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 product_qty_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 product_qty_condensed,
CASE WHEN 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 v
INNER JOIN vendor_booth_assignments vba
ON v.vendor_id = vba.vendor_id
ORDER BY v.vendor_name, vba.market_date;

/* SECTION 3 */

-- AGGREGATE
/* 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(vendor_id) AS booth_rental_count
FROM vendor_booth_assignments
GROUP BY vendor_id;



Expand All @@ -63,7 +92,22 @@ sticker to everyone who has ever spent more than $2000 at the market. Write a qu
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
customer.customer_id,
customer.customer_first_name,
customer.customer_last_name,
SUM(customer_purchases.quantity * customer_purchases.cost_to_customer_per_qty) AS total_spent
FROM customer
INNER JOIN customer_purchases
ON customer.customer_id = customer_purchases.customer_id
GROUP BY
customer.customer_id,
customer.customer_first_name,
customer.customer_last_name
HAVING SUM(customer_purchases.quantity * customer_purchases.cost_to_customer_per_qty) > 2000
ORDER BY
customer.customer_last_name,
customer.customer_first_name;


--Temp Table
Expand All @@ -77,7 +121,17 @@ When inserting the new vendor, you need to appropriately align the columns to be
-> To insert the new row use VALUES, specifying the value you want for each column:
VALUES(col1,col2,col3,col4,col5)
*/

CREATE TABLE temp.new_vendor AS
SELECT *
FROM vendor;
INSERT INTO temp.new_vendor
VALUES (
10,
'Thomas Superfood Store',
'Fresh Focused',
'Thomas',
'Rosenthal'
);


-- Date
Expand Down
140 changes: 134 additions & 6 deletions 02_activities/assignments/Cohort_8/assignment2.sql
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@@ -1,5 +1,5 @@
/* ASSIGNMENT 2 */
/* SECTION 2 */
/* SECTION 2 */

-- COALESCE
/* 1. Our favourite manager wants a detailed long list of products, but is afraid of tables!
Expand All @@ -21,7 +21,16 @@ 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 *
--identifying NULL
FROM product
WHERE product_name IS NULL
OR product_size IS NULL
OR product_qty_type IS NULL;
--COALESCE and concatenate
SELECT
COALESCE(product_name, '')|| ', ' || COALESCE(product_size, '')|| ' (' || COALESCE(product_qty_type, 'unit') || ')'
FROM product


--Windowed Functions
Expand All @@ -33,17 +42,44 @@ You can either display all rows in the customer_purchases table, with the counte
each new market date for each customer, or select only the unique market dates per customer
(without purchase details) and number those visits.
HINT: One of these approaches uses ROW_NUMBER() and one uses DENSE_RANK(). */
SELECT*

FROM(
SELECT
customer_id,
market_date,
dense_rank() OVER(PARTITION BY customer_id ORDER BY market_date) as number_visits
FROM customer_purchases
)


/* 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. */

SELECT
customer_id,
market_date

FROM(
SELECT
customer_id,
market_date,
ROW_NUMBER() OVER(PARTITION BY customer_id ORDER BY market_date DESC) as number_visits
FROM customer_purchases
)
WHERE number_visits = 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,
vendor_id,
market_date,
COUNT(*) OVER(PARTITION BY customer_id, product_id) as customer_purchase_count
FROM customer_purchases



Expand All @@ -59,11 +95,22 @@ 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,
TRIM(
SUBSTR(
product_name,
INSTR(product_name, '-') + 1
)
) AS product_description
FROM product;


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


SELECT *
FROM product
WHERE product_size REGEXP '[0-9]';

-- UNION
/* 1. Using a UNION, write a query that displays the market dates with the highest and lowest total sales.
Expand All @@ -75,6 +122,41 @@ 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 date_sales AS (
-- Create a CTE/temp table to find sales values grouped dates
SELECT
market_date,
SUM(quantity * cost_to_customer_per_qty) AS total_sales
FROM customer_purchases
GROUP BY market_date
),
ranked_dates AS (
/*Create anocther CTE/Temp table with a rank windowed function on the previous QUERY
to create BEST DAY and WORST DAY*/

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 date_sales
)
--QUERY the second temp table twice, once for the BEST DAY,once for the WORST DAY, with a UNION
SELECT
market_date,
total_sales,
'BEST DAY' AS label
FROM ranked_dates
WHERE best_rank = 1

UNION

SELECT
market_date,
total_sales,
'WORST DAY' AS label
FROM ranked_dates
WHERE worst_rank = 1;



Expand All @@ -91,7 +173,37 @@ 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 distinct_vendor_products AS (

SELECT DISTINCT
vendor_id,
product_id,
original_price
FROM vendor_inventory
),
vendor_product_customer AS (

SELECT
dvp.vendor_id AS vendor_id,
dvp.product_id AS product_id,
dvp.original_price AS product_price,
cust.customer_id AS customer_id
FROM distinct_vendor_products dvp
CROSS JOIN customer cust
)
SELECT
ven.vendor_name,
prod.product_name,
SUM(5 * vpc.product_price) AS money_made_per_product
FROM vendor_product_customer vpc
JOIN vendor ven ON vpc.vendor_id = ven.vendor_id
JOIN product prod ON vpc.product_id = prod.product_id
GROUP BY
ven.vendor_name,
prod.product_name
ORDER BY
ven.vendor_name,
prod.product_name;

-- INSERT
/*1. Create a new table "product_units".
Expand All @@ -104,14 +216,30 @@ Name the timestamp column `snapshot_timestamp`. */
/*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). */


CREATE TABLE product_units AS
SELECT
product_id,
product_name,
product_size,
product_category_id,
product_qty_type,
CURRENT_TIMESTAMP AS snapshot_timestamp
FROM product
WHERE product_qty_type = 'unit';

-- 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.*/


DELETE FROM product_units2
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@monzchan monzchan Nov 28, 2025

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The table product_units2 does not exist. You previously created product_units.

--ROWID because all the rows shared the same snapshot_timestamp, realized this after delete all the rows without using ROWID
WHERE ROWID = (
SELECT ROWID
FROM product_units2
ORDER BY snapshot_timestamp ASC
LIMIT 1
);

-- UPDATE
/* 1.We want to add the current_quantity to the product_units table.
Expand Down
28 changes: 28 additions & 0 deletions 05_src/sql/SELECT.sql
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@@ -0,0 +1,28 @@
/* MODULE 2 */
/* SELECT */


/* 1. Select everything in the customer table */
SELECT *FROM customer;

/* 2. Use sql as a calculator */



/* 3. Add order by and limit clauses */
SELECT *FROM customer
ORDER BY customer_last_name DESC--ZA
LIMIT 5;--ONLY 5 ROWS


/* 4. Select multiple specific columns */
SELECT product_name,
product_size,
product_qty_type
FROM product;

/* 5. Add a static value in a column */
SELECT '2025' AS [this_year],
vendor_name,
vendor_type
FROM vendor;