It’s time to learn how to manage numeric data.
Let’s learn how to use `COUNT` and `SUM`.

🆕 🌟
📌COUNT: find the number of rows that match specified criteria
📌SUM: find the total sum of a numeric column

Query 1. Using COUNT
``SELECT COUNT(*) FROM address; SELECT COUNT(address2) FROM address;``

Query 2. Using COUNT with DISTINCT
``````SELECT COUNT(rental_duration) FROM film;
SELECT COUNT(DISTINCT rental_duration) FROM film;``````

Query 3. Using SUM
``````SELECT SUM(length) FROM film;
SELECT SUM(length)/60 AS entire_hour from film;``````

### # COUNT

Q1. Finding the number of all columns and address2 columns in the address table

SELECT COUNT (column name)

Do you remember how to view all the columns in a table? We have to use `SELECT * FROM table name`. If we want only the number of columns and not the data itself, we can use the new query `COUNT`.

Add `COUNT` after `SELECT`, and since we want to see the number of rows (data) in all the columns, remember to use an asterisk (`*`).

``````SELECT COUNT(*) from address;
``````

What if we wanted to see only the number of rows in a specific category? Then instead of an asterisk, we would write the name of the column in the parentheses. Let’s try address2:

``SELECT COUNT(address2) FROM address;``

`COUNT(*)` calculates the number of columns, including the null value. If you want to know the number of columns without the null value, you have to write a specific column name inside the parentheses.

Q2. Finding the number of all columns and Unique columns of rental duration in the film table

SELECT COUNT (DISTINCT column name)

If we want to see the number of unique values in a column, we can use the `DISTINCT` command that we learned in the last tutorial.

Let’s look at how we can use that with `COUNT`.

You can count the number of all rows in rental duration using `SELECT COUNT(rental_duration)`. But what if you want to know the number of unique values in the column? In the practice example we used in the last tutorial, we saw the values 3, 4, 5, 6, and 7 (hours). In total, there are five different rental duration periods.

You can find this ‘total’ if you use `DISTINCT` with `SELECT COUNT`.  So you can see the result of 5 by using this statement:

``SELECT COUNT(DISTINCT rental_duration) FROM film;``

If you don’t use the `DISTINCT` query, you will get the total count of the different rental duration periods. Take a look at the picture below for a comparison:

### # SUM

Q3. Finding the total length of all films in the film table

SELECT SUM (column name)

Now we will try finding the total length of all films by using `SUM`.

`SUM` has the same structure we used with `COUNT`. You can write `SUM` right after `SELECT` and the column name inside the parentheses like this:

``SELECT SUM(length) FROM film;``

You can see the number “115272” in the results, but we don’t know how many hours that equals, since the value showed is in minutes. Let’s divide its value into 60 to see it by hours and add a new column name using `AS`.

🔑 We can use the symbols (`+`), (`-`), (`*`), (` /`) in SQL when the data is numeric. So we can see the total length of all films by using `SUM(length)/60`. Just add `AS entire_hour` to rename the new column of the final result.

``SELECT SUM(length)/60 AS entire_hour FROM film;``

### # Practice Time

Your challenge today is to find the number of Unique Values of replacement_cost in the film table. It’s pretty simple, so let’s use what we learned so far to find the values!

We have to use `COUNT` and `DISTINCT` together to see the number of unique values. If we only use `DISTINCT` here, we have to count the result manually, so it is better to use `COUNT` to make things easier.

``SELECT COUNT(DISTINCT replacement_cost) FROM film;``

The result is 21 →, meaning we have 21 different values (in this case, 21 different prices for replacing the movies) in the specified column.