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188 changes: 177 additions & 11 deletions 02_activities/assignments/assignment2.sql
Original file line number Diff line number Diff line change
Expand Up @@ -19,8 +19,14 @@ HINT: keep the syntax the same, but edited the correct components with the strin
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
/* 1. Write a query that selects from the customer_purchases table and numbers each customer’s
Expand All @@ -32,18 +38,87 @@ 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(). */

--dense_rank
SELECT
customer_id,
market_date,
DENSE_RANK()
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;

--row_number
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. */


--Reverse-rank
SELECT
customer_id,
market_date,
DENSE_RANK()
OVER (
PARTITION BY customer_id
ORDER BY market_date DESC
) AS recent_visit_rank
FROM (
SELECT DISTINCT
customer_id,
market_date
FROM customer_purchases
) AS unique_visits;

--filter recent visit
WITH ranked_visits AS (
SELECT
customer_id,
market_date,
DENSE_RANK()
OVER (
PARTITION BY customer_id
ORDER BY market_date DESC
) AS recent_visit_rank
FROM (
SELECT DISTINCT
customer_id,
market_date
FROM customer_purchases
) AS unique_visits
)
SELECT
customer_id,
market_date
FROM ranked_visits
WHERE recent_visit_rank = 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
cp.*,
COUNT(*)
OVER (
PARTITION BY customer_id,
product_id
) AS times_purchased
FROM customer_purchases AS cp;

-- String manipulations
/* 1. Some product names in the product table have descriptions like "Jar" or "Organic".
Expand All @@ -57,10 +132,27 @@ 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 TRIM(
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
Expand All @@ -73,7 +165,32 @@ 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 daily_totals AS (
SELECT
market_date,
SUM(quantity * cost_to_customer_per_qty) AS total_sales
FROM customer_purchases
GROUP BY market_date
)
SELECT
'Highest' AS which,
market_date,
total_sales
FROM daily_totals
WHERE total_sales = (
SELECT MAX(total_sales)
FROM daily_totals
)
UNION
SELECT
'Lowest' AS which,
market_date,
total_sales
FROM daily_totals
WHERE total_sales = (
SELECT MIN(total_sales)
FROM daily_totals
);


/* SECTION 3 */
Expand All @@ -88,28 +205,62 @@ Remember, CROSS JOIN will explode your table rows, so CROSS JOIN should likely b
Think a bit about the row counts: how many distinct vendors, product names are there (x)?
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 potential_revenue
FROM vendor_inventory AS vi
JOIN vendor AS v ON vi.vendor_id = v.vendor_id
JOIN product AS p ON vi.product_id = p.product_id
CROSS JOIN customer AS 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
*,
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_qty_type,
snapshot_timestamp
)
VALUES (
999,
'Apple Pie',
'Individual',
'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_name = 'Apple Pie'
AND snapshot_timestamp < (
SELECT MAX(snapshot_timestamp)
FROM product_units
WHERE product_name = 'Apple Pie'
);

-- UPDATE
/* 1.We want to add the current_quantity to the product_units table.
Expand All @@ -129,5 +280,20 @@ Finally, make sure you have a WHERE statement to update the right row,
When you have all of these components, you can run the update statement. */


ALTER TABLE product_units
ADD current_quantity INT;


UPDATE product_units AS pu
SET current_quantity = COALESCE(
(
SELECT vi.quantity
FROM vendor_inventory AS vi
WHERE vi.product_id = pu.product_id
ORDER BY vi.market_date DESC
LIMIT 1
),
0
);


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