UofT-DSI | sql - homework 4

02_activities/homework/homework_4.sql

@@ -17,8 +17,9 @@ 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') || ')' AS product_details
+FROM product;
 
 --Windowed Functions
 /* 1. Write a query that selects from the customer_purchases table and numbers each customer’s  
@@ -30,16 +31,52 @@ 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. */
 
+-- Query to reverse the numbering of each customer's visit
+SELECT 
+    customer_id, 
+    market_date, 
+    ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY market_date DESC) AS visit_number
+FROM 
+    customer_purchases;
+
+-- Query to filter only the most recent visit of each customer using a subquery
+SELECT 
+    customer_id,
+    market_date AS 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 
+            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(product_id) OVER (PARTITION BY customer_id, product_id) AS purchases_count
+FROM 
+    customer_purchases;
 
 
 -- String manipulations
@@ -54,11 +91,24 @@ 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 RTRIM(LTRIM(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 REGEXP '[0-9]';
 
 -- UNION
 /* 1. Using a UNION, write a query that displays the market dates with the highest and lowest total sales.
@@ -70,6 +120,40 @@ 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_value
+    FROM 
+        customer_purchases
+    GROUP BY 
+        market_date
+),
+ranked_sales AS (
+    SELECT 
+        market_date,
+        total_sales_value,
+        RANK() OVER (ORDER BY total_sales_value DESC) AS best_day_rank,
+        RANK() OVER (ORDER BY total_sales_value ASC) AS worst_day_rank
+    FROM 
+        sales_by_date
+)
+SELECT 
+    'Best Day' AS type_of_day, 
+    market_date, 
+    total_sales_value
+FROM 
+    ranked_sales
+WHERE 
+    best_day_rank = 1
 
+UNION ALL
 
-
+SELECT 
+    'Worst Day' AS type_of_day, 
+    market_date, 
+    total_sales_value
+FROM 
+    ranked_sales
+WHERE 
+    worst_day_rank = 1;