15. Solutions: GROUP BY
Solutions: GROUP BY
- Which account (by name) placed the earliest order? Your solution should have the account name and the date of the order.
SELECT a.name, o.occurred_at
FROM accounts a
JOIN orders o
ON a.id = o.account_id
ORDER BY occurred_at
LIMIT 1;
- Find the total sales in usd for each account. You should include two columns - the total sales for each company's orders in usd and the company name.
SELECT a.name, SUM(total_amt_usd) total_sales
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.name;
- Via what channel did the most recent (latest) web_event occur, which account was associated with this web_event? Your query should return only three values - the date, channel, and account name.
SELECT w.occurred_at, w.channel, a.name
FROM web_events w
JOIN accounts a
ON w.account_id = a.id
ORDER BY w.occurred_at DESC
LIMIT 1;
- Find the total number of times each type of channel from the web_events was used. Your final table should have two columns - the channel and the number of times the channel was used.
SELECT w.channel, COUNT(*)
FROM web_events w
GROUP BY w.channel
- Who was the primary contact associated with the earliest web_event?
SELECT a.primary_poc
FROM web_events w
JOIN accounts a
ON a.id = w.account_id
ORDER BY w.occurred_at
LIMIT 1;
- What was the smallest order placed by each account in terms of total usd. Provide only two columns - the account name and the total usd. Order from smallest dollar amounts to largest.
SELECT a.name, MIN(total_amt_usd) smallest_order
FROM accounts a
JOIN orders o
ON a.id = o.account_id
GROUP BY a.name
ORDER BY smallest_order;
Sort of strange we have a bunch of orders with no dollars. We might want to look into those.
- Find the number of sales reps in each region. Your final table should have two columns - the region and the number of sales_reps. Order from fewest reps to most reps.
SELECT r.name, COUNT(*) num_reps
FROM region r
JOIN sales_reps s
ON r.id = s.region_id
GROUP BY r.name
ORDER BY num_reps;
Code
If you need a code on the https://github.com/udacity.