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2 changes: 1 addition & 1 deletion 02_activities/assignments/Assignment2.md
Original file line number Diff line number Diff line change
Expand Up @@ -54,7 +54,7 @@ The store wants to keep customer addresses. Propose two architectures for the CU
**HINT:** search type 1 vs type 2 slowly changing dimensions.

```
Your answer...
If we capture customer addresses with Type 1 SCD, the address is simply overwritten when it changes, easy to manage, but no history is kept. With Type 2 SCD, each new address is stored as a new row with dates, so we keep full history. Type 1 is good if we only care about the current address, while Type 2 is better if we need to track past addresses (e.g., old shipping locations).
```

***
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98 changes: 86 additions & 12 deletions 02_activities/assignments/assignment2.sql
Original file line number Diff line number Diff line change
@@ -1,4 +1,4 @@
/* ASSIGNMENT 2 */
/* ASSIGNMENT 2 - By Syeda Ali */
/* SECTION 2 */

-- COALESCE
Expand All @@ -20,7 +20,12 @@ 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') || ')'
AS product_description
FROM product;

--Windowed Functions
/* 1. Write a query that selects from the customer_purchases table and numbers each customer’s
Expand All @@ -32,18 +37,33 @@ 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,
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY market_date) AS visit_number
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 *
FROM (
SELECT customer_id,
market_date,
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY market_date DESC) AS visit_number
FROM customer_purchases
) 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 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 +77,17 @@ 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 description
FROM product;

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


SELECT product_id,
product_name,
product_size
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 @@ -73,8 +99,26 @@ 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 label
FROM ranked_sales
WHERE best_rank = 1

UNION

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

/* SECTION 3 */

Expand All @@ -89,27 +133,57 @@ 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). */


SELECT v.vendor_name,
p.product_name,
SUM(5 * vi.original_price) AS total_revenue
FROM vendor_inventory vi
JOIN vendor v
ON v.vendor_id = vi.vendor_id
JOIN product p
ON p.product_id = vi.product_id
CROSS JOIN customer c
GROUP BY v.vendor_name, p.product_name;

-- INSERT
/*1. Create a new table "product_units".
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 p.*,
CURRENT_TIMESTAMP AS snapshot_timestamp
FROM product p
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
SELECT *,
CURRENT_TIMESTAMP AS snapshot_timestamp
FROM product
WHERE product_name = 'Apple Pie';

-- 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_units
WHERE product_id = (
SELECT product_id
FROM product
WHERE product_name = 'Apple Pie')

AND snapshot_timestamp = (
SELECT MIN(snapshot_timestamp)
FROM product_units
WHERE product_id = (
SELECT product_id
FROM product
WHERE product_name = 'Apple Pie')
);

-- UPDATE
/* 1.We want to add the current_quantity to the product_units table.
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Please complete the last question

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The other looks very good!

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