Understanding JOINS , Sub Query and Aggregate Functions - sql

Question

In MYSQL , I am trying to understand the Aggregate Functions and am trying some examples in the northwind schema.
The table employees has the following description.
+-----------------+-------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------------+-------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| company | varchar(50) | YES | MUL | NULL | |
| last_name | varchar(50) | YES | MUL | NULL | |
| first_name | varchar(50) | YES | MUL | NULL | |
| email_address | varchar(50) | YES | | NULL | |
| job_title | varchar(50) | YES | | NULL | |
| business_phone | varchar(25) | YES | | NULL | |
| home_phone | varchar(25) | YES | | NULL | |
| mobile_phone | varchar(25) | YES | | NULL | |
| fax_number | varchar(25) | YES | | NULL | |
| address | longtext | YES | | NULL | |
| city | varchar(50) | YES | MUL | NULL | |
| state_province | varchar(50) | YES | MUL | NULL | |
| zip_postal_code | varchar(15) | YES | MUL | NULL | |
| country_region | varchar(50) | YES | | NULL | |
| web_page | longtext | YES | | NULL | |
| notes | longtext | YES | | NULL | |
| attachments | longblob | YES | | NULL | |
+-----------------+-------------+------+-----+---------+----------------+
Also the data in the table is
mysql> select city , first_name,last_name from employees;
+----------+------------+----------------+
| city | first_name | last_name |
+----------+------------+----------------+
| Seattle | Nancy | Freehafer |
| Bellevue | Andrew | Cencini |
| Redmond | Jan | Kotas |
| Kirkland | Mariya | Sergienko |
| Seattle | Steven | Thorpe |
| Redmond | Michael | Neipper |
| Seattle | Robert | Zare |
| Redmond | Laura | Giussani |
| Seattle | Anne | Hellung-Larsen |
+----------+------------+----------------+
I am trying to understand how can I find the Average number of people from different cities.
Till Now , I have
mysql> select city,count(city) from employees group by city;
+----------+-------------+
| city | count(city) |
+----------+-------------+
| Bellevue | 1 |
| Kirkland | 1 |
| Redmond | 3 |
| Seattle | 4 |
+----------+-------------+
Also I have
SELECT SUM(inner_count_city) from
(
SELECT city AS inner_city,
COUNT(*) AS inner_count_city
FROM employees
GROUP BY inner_city
) temp_table;
+-----------------------+
| SUM(inner_count_city) |
+-----------------------+
| 9 |
+-----------------------+
I am struggling to proceed forward do this due to the following reasons.
I am not able to do a AVG(COUNT(city)) - cant do two aggregate function
I am also not sure , how to divide it by the sum of count of cities ( = 9).
Not sure if I should use unions or joins , or subqueries.
I am trying to do something like
select city , inner_count_city / sum (inner_count_city) from ..

Answer 1

SQL only supports one aggregation at a time; for multiple aggregations you need multiple subqueries/CTEs. If you want the average number of people in the cities, then you are almost there:
SELECT AVG(inner_count_city * 1.0) from
FROM (SELECT city AS inner_city, COUNT(*) AS inner_count_city
FROM employees
GROUP BY inner_city
) c;
Note that SQL Server does integer arithmetic on integers. So the average of 1 and 2 is 1, not 1.5. Hence the * 1.0.
You can do this without a subquery as well:
SELECT COUNT(*) * 1.0 / COUNT(DISTINCT city)
FROM employees;

Answer 2

You can apply AVG aggregate function in an outer query:
select avg(cnt)
FROM (select count(city) as cnt
from employees
group by city) as t
To get the percentage of people per city you can use the following query:
select city, count(city) * 100.0 / total_count as cnt
from employees
cross join (select count(*) as total_count from employees) AS t
group by city, total_count