diff --git a/02_activities/assignments/Cohort_8/Logical_Model.png b/02_activities/assignments/Cohort_8/Logical_Model.png new file mode 100644 index 000000000..bcbc83263 Binary files /dev/null and b/02_activities/assignments/Cohort_8/Logical_Model.png differ diff --git a/02_activities/assignments/Cohort_8/assignment1.sql b/02_activities/assignments/Cohort_8/assignment1.sql index c992e3205..abb33935b 100644 --- a/02_activities/assignments/Cohort_8/assignment1.sql +++ b/02_activities/assignments/Cohort_8/assignment1.sql @@ -4,17 +4,24 @@ --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), @@ -23,10 +30,19 @@ 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 customer_id >= 8 + AND customer_id <= 10; -- option 2 - +SELECT + *, + quantity * cost_to_customer_per_qty AS price +FROM customer_purchases +WHERE customer_id BETWEEN 8 AND 10; --CASE @@ -35,6 +51,15 @@ Using the product table, write a query that outputs the product_id and product_n 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 LOWER(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. @@ -42,12 +67,37 @@ add a column to the previous query called pepper_flag that outputs a 1 if the pr contains the word “pepper” (regardless of capitalization), and otherwise outputs 0. */ +SELECT + product_id, + product_name, + CASE + WHEN LOWER(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 + v.vendor_id, + v.vendor_name, + v.vendor_type, + v.vendor_owner_first_name, + v.vendor_owner_last_name, + vba.booth_number, + vba.market_date +FROM vendor AS v +INNER JOIN vendor_booth_assignments AS vba + ON v.vendor_id = vba.vendor_id +ORDER BY v.vendor_name, vba.market_date; /* SECTION 3 */ @@ -57,6 +107,13 @@ vendor_id field they both have in common, and sorts the result by vendor_name, t at the farmer’s market by counting the vendor booth assignments per vendor_id. */ +SELECT + vendor_id, + COUNT(*) AS times_rented +FROM vendor_booth_assignments +GROUP BY vendor_id +ORDER BY times_rented DESC, 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 @@ -64,12 +121,25 @@ 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 AS c +JOIN customer_purchases AS 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 /* 1. Insert the original vendor table into a temp.new_vendor and then add a 10th vendor: Thomass Superfood Store, a Fresh Focused store, owned by Thomas Rosenthal + + + HINT: This is two total queries -- first create the table from the original, then insert the new 10th vendor. When inserting the new vendor, you need to appropriately align the columns to be inserted (there are five columns to be inserted, I've given you the details, but not the syntax) @@ -78,6 +148,26 @@ 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 TEMP TABLE new_vendor AS +SELECT * FROM vendor; + +INSERT INTO new_vendor ( + vendor_id, + vendor_name, + vendor_type, + vendor_owner_first_name, + vendor_owner_last_name +) +VALUES ( + 10, + 'Thomass Superfood Store', + 'Fresh Focused', + 'Thomas', + 'Rosenthal' +); + + -- Date @@ -87,6 +177,13 @@ HINT: you might need to search for strfrtime modifers sqlite on the web to know and year are! */ +SELECT + customer_id, + STRFTIME('%m', purchase_datetime) AS month, + STRFTIME('%Y', purchase_datetime) AS year +FROM customer_purchases; + + /* 2. Using the previous query as a base, determine how much money each customer spent in April 2022. Remember that money spent is quantity*cost_to_customer_per_qty. @@ -94,3 +191,11 @@ Remember that money spent is quantity*cost_to_customer_per_qty. HINTS: you will need to AGGREGATE, GROUP BY, and filter... but remember, STRFTIME returns a STRING for your WHERE statement!! */ +SELECT + customer_id, + SUM(quantity * cost_to_customer_per_qty) AS total_spent_apr_2022 +FROM customer_purchases +WHERE STRFTIME('%Y', purchase_datetime) = '2022' + AND STRFTIME('%m', purchase_datetime) = '04' +GROUP BY customer_id +ORDER BY total_spent_apr_2022 DESC, customer_id;