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Ordering
Eliminating duplicate results
Aggregation
Grouping
16. Group the rows and count them
HAVING: filtering and ordering groups
Let's practice

## Instruction

In the previous section, we've learnt how to count statistics for all rows. We'll now go on to study even more sophisticated statistics. Look at the following statement:

SELECT
customer_id,
count(*)
FROM orders
GROUP BY customer_id;


The new piece here is GROUP BY followed by a column name (customer_id). GROUP BY will group together all rows having the same value in the specified column.

In our example, all orders made by the same customer will be grouped together in one row. The function count(*) will then count all rows for the specific clients. As a result, we'll get a table where each customer_id will be shown together with the number of orders placed by that customer.

Take a look at the following table which illustrates the query:

order_id customer_id order_date ship_date total_sum customer_id count(*)
1 1 21-02-2014 22-02-2014 1009.00 1 3
2 1 25-02-2014 25-02-2014 2100.00
3 1 03-03-2014 03-03-2014 315.00
4 2 03-03-2014 04-03-2014 401.67 2 2
5 2 03-03-2014 07-03-2014 329.29
6 3 15-03-2014 15-03-2014 25349.68 3 1
7 4 19-03-2014 20-03-2014 2324.32 4 4
8 4 02-04-2014 02-04-2014 7542.21
9 4 05-04-2014 07-04-2014 123.23
10 4 05-04-2014 07-04-2014 425.33
11 5 06-04-2014 09-04-2014 2134.65 5 5
12 5 17-04-2014 19-04-2014 23.21
13 5 25-04-2014 26-04-2014 5423.23
14 5 29-04-2014 30-04-2014 4422.11
15 5 30-04-2014 30-04-2014 532.54

## Exercise

Find the number of employees in each department in the year 2013. Show the department name together with the number of employees.

### Stuck? Here's a hint!

Type:

SELECT
department,
count(*)
FROM employees
WHERE year = 2013
GROUP BY department;