Assignment two

02_activities/assignments/Assignment2.md

@@ -54,7 +54,9 @@ 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...
+type 1 slowly changing dimension is the one where old data is overwritten with the new data. There is no record or history of old data that's why the reports generated with this architecture shows NULL or wrong values sometimes.
+type 2 slowly changing dimesion architecture keeps the historical data and enables tracking changes in dimension data over time by adding a new row for each change in the table.
+The two images of customer table depicts the type 1 and type 2 SCD architecture.
 ```
 
 ***

02_activities/assignments/assignment2.sql

@@ -20,6 +20,10 @@ 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
@@ -32,18 +36,49 @@ 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 (
+    SELECT DISTINCT customer_id, market_date
+    FROM customer_purchases
+) AS unique_visits
+ORDER BY customer_id, visit_number;
 
 
 /* 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 visit_number
+    FROM (
+        SELECT DISTINCT customer_id, market_date
+        FROM customer_purchases
+    ) AS unique_visits
+) x
+WHERE visit_number = 1
+ORDER BY customer_id;
 
 /* 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,
+    quantity,
+    cost_to_customer_per_qty,
+    COUNT(*) OVER(
+        PARTITION BY customer_id, product_id
+    ) AS purchase_count
+FROM customer_purchases
+ORDER BY customer_id, product_id, market_date;
 
 -- String manipulations
 /* 1. Some product names in the product table have descriptions like "Jar" or "Organic". 
@@ -57,11 +92,28 @@ 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_id, product_name,
+    CASE 
+        WHEN INSTR(product_name, '-') > 0 THEN 
+            TRIM(SUBSTR(product_name, INSTR(product_name, '-') + 1))
+        ELSE 
+            NULL
+    END AS description, 
+	product_size, product_qty_type
+FROM product;
 
 /* 2. Filter the query to show any product_size value that contain a number with REGEXP. */
 
-
+SELECT  product_id, product_name,
+    CASE 
+        WHEN INSTR(product_name, '-') > 0 THEN 
+            TRIM(SUBSTR(product_name, INSTR(product_name, '-') + 1))
+        ELSE 
+            NULL
+    END AS description, 
+	product_size, product_qty_type
+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.
@@ -72,9 +124,75 @@ HINT: There are a possibly a few ways to do this query, but if you're struggling
 "best day" and "worst day"; 
 3) Query the second temp table twice, once for the best day, once for the worst day, 
 with a UNION binding them. */
-
-
-
+-- Answer is in two alternate ways RESULT QUERY 1 and RESULT QUERY 2
+
+-- *** RESULT QUERY 1: ***
+DROP VIEW IF EXISTS vendor_daily_sales;
+CREATE VIEW IF NOT EXISTS vendor_daily_sales AS
+	SELECT md.market_date, market_day, market_year, vendor_name
+	,SUM(quantity*cost_to_customer_per_qty) as sales
+	FROM market_date_info md
+	INNER JOIN customer_purchases cp
+		ON md.market_date = cp.market_date
+	INNER JOIN vendor v										-- This JOIN is optional in this query 
+		ON v.vendor_id = cp.vendor_id
+    GROUP BY cp.market_date, v.vendor_id;
+
+	-- Execute following query after running the above query (part of RESULT QUERY 1).
+
+WITH daily_sales AS (
+    SELECT market_date, SUM(sales) AS total_sales FROM vendor_daily_sales  GROUP BY market_date),
+total_sales_days 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 daily_sales)
+
+SELECT market_date, total_sales, 'Best Day' AS max_min_sales FROM total_sales_days WHERE best_rank = 1
+UNION
+SELECT market_date, total_sales, 'Worst Day' AS max_min_sales FROM total_sales_days WHERE worst_rank = 1 
+ORDER BY max_min_sales;
+
+-- ***END*** RESULT QUERY 1
+
+-- *** RESULT QUERY 2: *** (You can run whole RESULT QUERY 2 altogether)
+
+DROP VIEW IF EXISTS vendor_daily_sales;
+CREATE VIEW IF NOT EXISTS vendor_daily_sales AS
+	SELECT md.market_date, market_day, market_year, vendor_name
+	,SUM(quantity*cost_to_customer_per_qty) as sales
+	FROM market_date_info md
+	INNER JOIN customer_purchases cp
+		ON md.market_date = cp.market_date
+	INNER JOIN vendor v										-- This JOIN is optional in this query 
+		ON v.vendor_id = cp.vendor_id
+    GROUP BY cp.market_date, v.vendor_id;
+
+SELECT market_date, total_sales, 'Best Day' AS max_min_sales 
+FROM (
+	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 (
+		SELECT market_date, SUM(sales) AS total_sales FROM vendor_daily_sales GROUP BY market_date
+	) as daily_sales
+) as total_sales_days WHERE best_rank = 1
+
+UNION
+
+SELECT market_date, total_sales, 'Worst Day' AS max_min_sales 
+FROM (
+	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 (
+		SELECT market_date, SUM(sales) AS total_sales FROM vendor_daily_sales GROUP BY market_date
+	) as daily_sales
+) as total_sales_days WHERE worst_rank = 1
+
+ORDER BY max_min_sales;
+
+-- ***END*** RESULT QUERY 2
 
 /* SECTION 3 */
 
@@ -89,6 +207,48 @@ 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).  */
 
+-- Not sure if we have to calculate the profit as well. We can also calculate profit by uncommenting lines in the following query.
+
+WITH vendor_products AS (
+    SELECT 
+        v.vendor_id,
+        v.vendor_name,
+        p.product_id,
+        p.product_name,
+		--cp.cost_to_customer_per_qty,
+        vi.original_price
+    FROM vendor_inventory vi
+    INNER JOIN vendor v ON vi.vendor_id = v.vendor_id
+    INNER JOIN product p ON vi.product_id = p.product_id
+	INNER JOIN customer_purchases cp ON vi.vendor_id = cp.vendor_id
+),
+customers AS (
+    SELECT 
+        customer_id
+    FROM customer
+),
+expected_sales AS (
+    SELECT 
+        vp.vendor_id,
+        vp.vendor_name,
+        vp.product_id,
+        vp.product_name,
+		--vp.cost_to_customer_per_qty * 5 AS revenue_per_customer,
+        vp.original_price * 5 AS cost_per_customer
+    FROM vendor_products vp
+    CROSS JOIN customers c
+)
+
+SELECT 
+    vendor_id,
+    vendor_name,
+    product_name,
+    --SUM(revenue_per_customer) AS total_revenue,
+	SUM(cost_per_customer) AS total_expected_revenue--,
+    --SUM(revenue_per_customer - cost_per_customer) AS total_profit
+FROM expected_sales
+GROUP BY vendor_id, vendor_name, product_name
+ORDER BY vendor_name, product_name;
 
 
 -- INSERT
@@ -97,19 +257,25 @@ 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`. */
 
+DROP TABLE IF EXISTS product_units;
+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 (6666,  'Strawberry Pie', '9 inch', 3, '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.*/
 
-
+DELETE FROM product_units WHERE product_id = 6666;
 
 -- UPDATE
 /* 1.We want to add the current_quantity to the product_units table. 
@@ -128,6 +294,19 @@ 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 COLUMN current_quantity INT;
+
+SELECT product_id, COALESCE(MAX(quantity), 0) AS latest_quantity
+FROM vendor_inventory
+GROUP BY product_id;
+
+UPDATE product_units
+SET current_quantity = (
+    SELECT COALESCE(MAX(vi.quantity), 0) FROM vendor_inventory vi
+    WHERE vi.product_id = product_units.product_id
+)
+WHERE EXISTS (
+    SELECT 1 FROM vendor_inventory vi
+    WHERE vi.product_id = product_units.product_id
+);