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Introduction
Nested CTEs
Summary

## Instruction

Alright, let's get started! We're going to tackle nested CTEs.

If you take a look into any database documentation, you will find out that you cannot put one CTE inside the parentheses of another CTE. Luckily, we can still use a simple technique to create nested CTEs. Take a look at the query:

WITH total_salesman AS (
SELECT
s.id AS s_id,
c.id AS c_id,
SUM(distance) AS sum_kilometers
FROM salesman s
JOIN daily_sales ds
ON s.id=ds.salesman_id
JOIN city c
ON s.city_id=c.id
GROUP BY s.id, c.id
),

max_salesman AS (
SELECT
c_id,
MAX(sum_kilometers) AS max_kilometers
FROM total_salesman
GROUP BY c_id
)

SELECT
AVG(max_kilometers)
FROM max_salesman;

The trick is that the second query, max_salesman, uses the first CTE, total_salesman, in the FROM clause (FROM total_salesman). That means that once we define the first CTE, we can freely use it in subsequent CTEs.

Now, what does that query do? In the first CTE, we find the total number of kilometers driven in each city by each salesman (on all days). In the second CTE, we take these sums and choose the salesman with the maximal distance in each city. Finally, in the outer query, we find the average maximal distance across all cities.

As you know, it's impossible to use one aggregate function inside another. Nested CTEs are a simple way to get rid of that restriction.

## Exercise

Now it's your turn! In this part, we will clearly state each step you need to take so that you don't get confused.

First, find daily sums of amount_earned in each city. Then, find the average daily amount for all cities for all days. Finally, show the id and name of each city plus the number of daily sums that exceeded the average daily amount for each given city.

### Stuck? Here's a hint!

SUM amounts earned in the first CTE, then find the average in the second CTE. In the outer query, use COUNT and make use of the results from both CTEs in the WHERE clause:

WHERE sum_amount > avg_amount 