diff --git a/02_activities/assignments/DC_Cohort/Assignment1.md b/02_activities/assignments/DC_Cohort/Assignment1.md index f78778f5b..c4335425b 100644 --- a/02_activities/assignments/DC_Cohort/Assignment1.md +++ b/02_activities/assignments/DC_Cohort/Assignment1.md @@ -207,3 +207,6 @@ Consider, for example, concepts of fariness, inequality, social structures, marg ``` Your thoughts... ``` +It’s sad that after being educated for many years, we’ve come to understand that people can’t be simply divided into “good” and “bad,” yet on the internet, our choices are often reduced to binary options like “Yes/No,” “True/False,” or “If/Else.” Our nuanced, unspeakable reasons get cut off when faced with such rigid conditions. +I once heard that people with the last name Null have experienced countless system breakdowns, to the point that some had to change their names — a striking example of how human identity can be distorted by data logic. +Regarding the article, I don’t want to judge its specific case, but it feels cold that we’re constantly trying to force every human situation into a perfect logical match, even when the settings are absurd. Such database-driven criteria could also become a terrifying tool of control — one that can be used to unlawfully restrict human rights. \ No newline at end of file diff --git a/02_activities/assignments/DC_Cohort/assignment1.sql b/02_activities/assignments/DC_Cohort/assignment1.sql index c992e3205..7d8abd04e 100644 --- a/02_activities/assignments/DC_Cohort/assignment1.sql +++ b/02_activities/assignments/DC_Cohort/assignment1.sql @@ -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=4 OR product_id=9; /*2. Write a query that returns all customer purchases and a new calculated column 'price' (quantity * cost_to_customer_per_qty), @@ -23,10 +29,16 @@ 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 @@ -34,28 +46,42 @@ filtered by customer IDs between 8 and 10 (inclusive) using either: 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 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. 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 prod_qty_type_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_booth_assignments +INNER JOIN vendor + ON vendor.vendor_id = vendor_booth_assignments.vendor_id +ORDER BY vendor_name, 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 times +FROM vendor_booth_assignments +GROUP BY vendor_id; /* 2. The Farmer’s Market Customer Appreciation Committee wants to give a bumper @@ -63,8 +89,17 @@ 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 + c.customer_id, + c.customer_first_name, + c.customer_last_name, + SUM(cp.quantity * cp.cost_to_customer_per_qty) AS total_spend +FROM customer AS c +INNER 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: @@ -77,7 +112,14 @@ 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) */ - + DROP TABLE IF EXISTS temp.new_vendor; + CREATE TABLE temp.new_vendor AS +SELECT * +FROM vendor; +INSERT INTO temp.new_vendor +VALUES (10, 'Thomass Superfood Store', 'Fresh Focused', 'Thomas', 'Rosenthal'); +SELECT * +FROM temp.new_vendor; -- Date diff --git a/02_activities/assignments/DC_Cohort/section 1.png b/02_activities/assignments/DC_Cohort/section 1.png new file mode 100644 index 000000000..e31a026ff Binary files /dev/null and b/02_activities/assignments/DC_Cohort/section 1.png differ