Aggregations.sql

-- QUESTIONS: SUM
-- 1. Find the total amount of poster_qty paper ordered in the orders table.
SELECT SUM(poster_qty) total_poster_qty
FROM orders;

-- 2. Find the total amount of standard_qty paper ordered in the orders table.
SELECT SUM(standard_qty) total_standard_qty
FROM orders;

-- 3. Find the total dollar amount of sales using the total_amt_usd in the orders table.
SELECT SUM(total_amt_usd) total_dollar_Amt
FROM orders;

-- 4. Find the total amount spent on standard_amt_usd and gloss_amt_usd paper for each order in the orders table. This should give a dollar amount for each order in the table
SELECT	standard_amt_usd + gloss_amt_usd total_amt_spent
FROM orders;

-- 5. Find the standard_amt_usd per unit of standard_qty paper. Your solution should use both an aggregation and a mathematical operator.
SELECT SUM(standard_amt_usd)/SUM(standard_qty) total_amt_per_unit
FROM orders;


-- QUESTIONS: MIN, MAX, & AVG
-- 1. When was the earliest order ever placed? 
SELECT MIN(occurred_at) Earliest_order_date
FROM orders;

-- 2. Try performing the same query as in question 1 without using an aggregation function.
SELECT occurred_at Earliest_order_date
FROM orders
ORDER BY occurred_at ASC
LIMIT 1;

-- 3. When did the most recent (latest) web_event occur?
SELECT MAX(occurred_at) latest_web_event_date
FROM web_events;

-- 4. Try to perform the result of the previous query without using an aggregation function
SELECT occurred_at latest_web_event_date
FROM web_events
ORDER BY occurred_at DESC
LIMIT 1;

-- 5. Find the mean (AVERAGE) amount spent per order on each paper type, as well as the mean amount of each paper type purchased per order. Your final answer should have 6 values - one for each paper type for the average number of sales, as well as the average amount.
SELECT	AVG(standard_amt_usd) Avg_standard_amt,
		AVG(gloss_amt_usd) Avg_gloss_amt,
       	AVG(poster_amt_usd) Avg_poster_amt,
        AVG(standard_qty) Avg_standard_qty,
		AVG(gloss_qty) Avg_gloss_qty,
       	AVG(poster_qty) Avg_poster_qty
FROM orders;


-- QUESTIONS: GROUP BY
-- 1. 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 o.account_id = a.id
ORDER BY o.occurred_at ASC
LIMIT 1;

-- 2. 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(o.total_amt_usd) total_sales_usd
FROM accounts a
JOIN orders o 
ON o.account_id = a.id
GROUP BY a.name;

-- 3. 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 a.id = w.account_id
ORDER BY w.occurred_at DESC
LIMIT 1;

-- 4. 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 channel, COUNT(channel)
FROM web_events
GROUP BY channel;

-- 5. Who was the primary contact associated with the earliest web_event?
SELECT w.occurred_at, a.primary_poc
FROM web_events w
JOIN accounts a
ON a.id = w.account_id
ORDER BY w.occurred_at ASC
LIMIT 1;

-- 6. 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(o.total_amt_usd) lowest_order
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.name
ORDER BY lowest_order ASC;

-- 7. 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 region, COUNT(s.name) no_of_sales_reps 
FROM sales_reps s
JOIN region r
ON r.id = s.region_id
GROUP BY r.name
ORDER BY no_of_sales_reps;

-- 8. For each account, determine the average amount of each type of paper they purchased across their orders. Your result should have four columns - one for the account name and one for the average quantity purchased for each of the paper types for each account.
SELECT a.name account_name, ROUND(AVG(standard_qty)) Avg_standard_qty, ROUND(AVG(gloss_qty)) Avg_gloss_qty, ROUND(AVG(poster_qty)) Avg_poster_qty
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.name
ORDER BY a.name;

-- 9. For each account, determine the average amount spent per order on each paper type. Your result should have four columns - one for the account name and one for the average amount spent on each paper type.
SELECT a.name account_name, ROUND(AVG(standard_amt_usd)) Avg_standard_amt, ROUND(AVG(gloss_amt_usd)) Avg_gloss_amt, ROUND(AVG(poster_amt_usd)) Avg_poster_amt
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.name
ORDER BY a.name;
                                      
 -- 10. Determine the number of times a particular channel was used in the web_events table for each sales rep. Your final table should have three columns - the name of the sales rep, the channel, and the number of occurrences. Order your table with the highest number of occurrences first
SELECT s.name sales_rep_name, w.channel, COUNT(*) num_of_events
FROM web_events w
JOIN accounts a 
ON a.id = w.account_id
JOIN sales_reps s
ON s.id = a.sales_rep_id
GROUP BY s.name, w.channel
ORDER BY num_of_events DESC;

-- 11. Determine the number of times a particular channel was used in the web_events table for each region. Your final table should have three columns - the region name, the channel, and the number of occurrences. Order your table with the highest number of occurrences first.                          
SELECT r.name region, w.channel, COUNT(*) num_of_events
FROM web_events w
JOIN accounts a 
ON a.id = w.account_id
JOIN sales_reps s
ON s.id = a.sales_rep_id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name, w.channel
ORDER BY num_of_events DESC;


-- QUESTIONS: Using DISTINCT
-- 1. Use DISTINCT to test if there are any accounts associated with more than one region.
SELECT DISTINCT id, name
FROM accounts;

-- 2. Have any sales reps worked on more than one account?
SELECT s.id, s.name, COUNT(a.sales_rep_id)
FROM accounts a
JOIN sales_reps s
ON s.id = a.sales_rep_id
GROUP BY s.id, s.name
ORDER BY COUNT(a.sales_rep_id);


-- QUESTIONS: HAVING
-- 1.  How many of the sales reps have more than 5 accounts that they manage?
SELECT s.id, s.name, COUNT(*) num_of_accounts
FROM accounts a
JOIN sales_reps s
ON s.id = a.sales_rep_id
GROUP BY s.id, s.name
HAVING COUNT(*) > 5
ORDER BY num_of_accounts DESC;

-- 2. How many accounts have more than 20 orders
SELECT a.id, a.name account_name, COUNT(*) num_of_orders
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id,a.name
HAVING COUNT(*) > 20
ORDER BY num_of_orders DESC;

-- 3. Which account has the most orders?
SELECT a.id, a.name account_name, COUNT(*) num_of_orders
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY num_of_orders DESC
LIMIT 1;

-- 4. How many accounts spent more than 30,000 usd total across all orders?
SELECT a.id, a.name account_name, COUNT(*) num_of_orders, SUM(total_amt_usd) total_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
HAVING SUM(total_amt_usd) > 30000
ORDER BY total_spent DESC;

-- 5. Which accounts spent less than 1,000 usd total across all orders?
SELECT a.id, a.name account_name, SUM(total_amt_usd) total_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
HAVING SUM(total_amt_usd) < 1000
ORDER BY total_spent DESC;

-- 6. Which account has spent the most with us
SELECT a.id, a.name account_name, SUM(total_amt_usd) total_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY total_spent DESC
LIMIT 1;

-- 7. Which account has spent the least with us?
SELECT a.id, a.name account_name, SUM(total_amt_usd) total_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY total_spent
LIMIT 1;

-- 8. Which accounts used facebook as a channel to contact customers more than 6 times?
SELECT a.id, a.name account_name, w.channel, COUNT(w.channel)
FROM web_events w
JOIN accounts a
ON a.id = w.account_id
AND w.channel = 'facebook'
GROUP BY a.id, a.name, w.channel
HAVING COUNT(w.channel) > 6
ORDER BY COUNT(w.channel) DESC;

-- 9. Which account used facebook most as a channel?
SELECT a.id, a.name account_name, w.channel, COUNT(w.channel)
FROM web_events w
JOIN accounts a
ON a.id = w.account_id
AND w.channel = 'facebook'
GROUP BY a.id, a.name, w.channel
HAVING COUNT(w.channel) > 6
ORDER BY COUNT(w.channel) DESC
LIMIT 1;

-- 10. Which channel was most frequently used by most accounts?
SELECT a.id, a.name account_name, w.channel, COUNT(w.channel) num_of_use
FROM web_events w
JOIN accounts a
ON a.id = w.account_id
GROUP BY a.id, a.name, w.channel
ORDER BY COUNT(w.channel) DESC
LIMIT 10;


-- QUESTIONS: DATE Functions
-- 1. Find the sales in terms of total dollars for all orders in each year, ordered from greatest to least.
SELECT DATE_PART('year',occurred_at), SUM(total_amt_usd) sales
FROM orders
GROUP BY 1
ORDER BY 2 DESC;

SELECT year(occurred_at), SUM(total_amt_usd) sales -- year is used in mysql while DATE_PART is used in Postgresql
FROM orders
GROUP BY 1
ORDER BY 2 DESC;

-- 2. Which month did Parch & Posey have the greatest sales in terms of total dollars? Are all months evenly represented by the dataset?
SELECT DATE_PART('month',occurred_at), SUM(total_amt_usd) sales -- syntax for Postgresql
FROM orders
WHERE occurred_at BETWEEN '2014-01-01' AND '2017-01-01'
GROUP BY 1
ORDER BY 2 DESC;

SELECT Month(occurred_at), SUM(total_amt_usd) sales -- syntax for MySQL
FROM orders
WHERE occurred_at BETWEEN '2014-01-01' AND '2017-01-01'
GROUP BY 1
ORDER BY 2 DESC;

-- 3. Which year did Parch & Posey have the greatest sales in terms of total number of orders? Are all years evenly represented by the dataset?
SELECT DATE_PART('year',occurred_at) , SUM(total) orders -- syntax for Postgresql
FROM orders
GROUP BY 1
ORDER BY 2 DESC;

SELECT YEAR(occurred_at) , SUM(total) orders -- syntax for MySql
FROM orders
GROUP BY 1
ORDER BY 2 DESC;

-- 4. Which month did Parch & Posey have the greatest sales in terms of total number of orders? Are all months evenly represented by the dataset?
SELECT DATE_PART('month',occurred_at) , SUM(total) orders  -- syntax for Postgresql
FROM orders
WHERE occurred_at BETWEEN '2014-01-01' AND '2017-01-01'
GROUP BY 1
ORDER BY 2 DESC;

SELECT MONTH(occurred_at) , SUM(total) orders -- syntax for MySql
FROM orders
WHERE occurred_at BETWEEN '2014-01-01' AND '2017-01-01'
GROUP BY 1
ORDER BY 2 DESC;

-- 5. In which month of which year did Walmart spend the most on gloss paper in terms of dollars?
SELECT DATE_TRUNC('month',o.occurred_at) order_date, SUM(o.gloss_amt_usd) gloss_amt_usd  -- syntax for Postgresql
FROM orders o
JOIN accounts a
ON a.id = o.account_id
WHERE a.name = 'Walmart'
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1;

SELECT EXTRACT(YEAR_MONTH FROM o.occurred_at) order_date, SUM(o.gloss_amt_usd) gloss_amt_usd  -- syntax for MySql
FROM orders o
JOIN accounts a
ON a.id = o.account_id
WHERE a.name = 'Walmart'
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1;


-- QUESTIONS: CASE
-- 1. Write a query to display for each order, the account ID, total amount of the order, and the level of the order - ‘Large’ or ’Small’ - depending on if the order is $3000 or more, or smaller than $3000
SELECT account_id, total_amt_usd,
   CASE WHEN total_amt_usd >= 3000 THEN 'Large'
   ELSE 'Small' END AS order_level
FROM orders;

-- 2. Write a query to display the number of orders in each of three categories, based on the total number of items in each order. The three categories are: 'At Least 2000', 'Between 1000 and 2000' and 'Less than 1000'.
SELECT CASE 
   		WHEN total >= 2000 THEN 'At least 2000'
   		WHEN total >= 1000 AND total < 2000 THEN 'Between 1000 and 2000'
   		ELSE 'Less than 1000' END AS num_of_orders,
COUNT(*) AS order_count
FROM orders
GROUP BY 1;

-- 3. We would like to understand 3 different levels of customers based on the amount associated with their purchases. The top level includes anyone with a Lifetime Value (total sales of all orders) greater than 200,000 usd. The second level is between 200,000 and 100,000 usd. The lowest level is anyone under 100,000 usd. Provide a table that includes the level associated with each account. You should provide the account name, the total sales of all orders for the customer, and the level. Order with the top spending customers listed first.
SELECT a.name account_name, SUM(total_amt_usd),
   CASE  
   		WHEN SUM(total_amt_usd) > 200000 THEN 'Top'
   		WHEN SUM(total_amt_usd) >= 100000 THEN 'Middle'
   		ELSE 'Low' END AS Customers_Level
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.name
ORDER BY 2 DESC;

-- 4. We would now like to perform a similar calculation to the first, but we want to obtain the total amount spent by customers only in 2016 and 2017. Keep the same levels as in the previous question. Order with the top spending customers listed first.
SELECT a.name account_name, SUM(total_amt_usd),
   CASE  
   		WHEN SUM(total_amt_usd) > 200000 THEN 'Top'
   		WHEN SUM(total_amt_usd) >= 100000 THEN 'Middle'
   		ELSE 'Low' END AS Customers_Level
FROM orders o
JOIN accounts a
ON a.id = o.account_id
AND o.occurred_at > '2015-12-31'
GROUP BY a.name
ORDER BY 2 DESC;

-- 5. We would like to identify top performing sales reps, which are sales reps associated with more than 200 orders. Create a table with the sales rep name, the total number of orders, and a column with top or not depending on if they have more than 200 orders. Place the top sales people first in your final table.
SELECT s.name sales_rep, COUNT(total) total_orders,
   CASE  
   		WHEN COUNT(total) > 200 THEN 'Top'
		ELSE 'Not' END AS Sales_rep_Level
FROM orders o
JOIN accounts a
ON a.id = o.account_id
JOIN sales_reps s
ON s.id = a.sales_rep_id
GROUP BY s.name
ORDER BY 2 DESC;

-- 6. The previous didn't account for the middle, nor the dollar amount associated with the sales. Management decides they want to see these characteristics represented as well. We would like to identify top performing sales reps, which are sales reps associated with more than 200 orders or more than 750000 in total sales. The middle group has any rep with more than 150 orders or 500000 in sales. Create a table with the sales rep name, the total number of orders, total sales across all orders, and a column with top, middle, or low depending on this criteria. Place the top sales people based on dollar amount of sales first in your final table. You might see a few upset sales people by this criteria!
SELECT s.name sales_rep, COUNT(*) total_orders, SUM(o.total_amt_usd) total_spent,
   CASE  
   		WHEN COUNT(*) > 200 OR SUM(o.total_amt_usd) > 750000 THEN 'Top'
        WHEN COUNT(*) > 150 OR SUM(o.total_amt_usd) > 500000 THEN 'Middle'
		ELSE 'Low' END AS sales_rep_level
FROM orders o
JOIN accounts a
ON a.id = o.account_id
JOIN sales_reps s
ON s.id = a.sales_rep_id
GROUP BY s.name
ORDER BY 3 DESC;