MySQL Join 3 Tables with Count - mysql

Question

I have the following tables:
users
id
username
first_name
email
students
id
user_id
payments
id
student_id
is_paid
date
And I have the following query:
SELECT users.username, users.first_name, users.email,
COUNT(payments.id) as pending_payments
FROM users
LEFT JOIN students ON users.id = students.user_id
LEFT JOIN payments ON payments.student_id = students.id AND payments.is_paid = 0
WHERE users.username LIKE 'testuser';
When the username exists and there are not pending payments, it returns:
| username | first_name | email | pending_payments |
| testuser | test | testuser#example.com | 0 |
When the username exists and there are 2 pending payments, it returns:
| username | first_name | email | pending_payments |
| testuser2| test2 | testuser#2example.com | 2 |
But when the username does not exists, it returns:
| username | first_name | email | pending_payments |
| NULL | NULL | NULL | 0 |
Instead, when the username does not exists, I want as result:
Empty set.
How can I modify my query to get the expected results?
(keep the behavior when user exists, but returns Empty set when does not exists, instead of a row will NULL values).
See SQLfiddle: http://sqlfiddle.com/#!9/aab424/11
EDIT
A working solution was provided:
http://sqlfiddle.com/#!9/aab424/72

Answer 1

Aggregate functions like COUNT(*) always generate one row if no GROUP BY clause is used. So use a GROUP BY clause, if you want an empty result:
SELECT users.username, users.first_name, users.email,
COUNT(payments.id) as pending_payments
FROM users
LEFT JOIN students ON users.id = students.user_id
LEFT JOIN payments ON payments.student_id = students.id AND payments.is_paid = 0
WHERE users.username LIKE 'testuser3'
GROUP BY users.username, users.first_name, users.email;
Note that you will get one row per user with that username. If username is not UNIQUE and you want to get one row for all users with the same username, you can use the HAVING clause instead of GROUP BY to "remove" the "empty" row.
SELECT users.username, COUNT(payments.id) as pending_payments
FROM users
LEFT JOIN students ON users.id = students.user_id
LEFT JOIN payments ON payments.student_id = students.id AND payments.is_paid = 0
WHERE users.username LIKE 'testuser3'
HAVING username IS NOT NULL;
SQLFiddle
This has nothing to do with the LEFT or any JOIN. It is just how aggregate functions work (in MySQL), and you will see the same behavior with a simple example without joins like
SELECT username, COUNT(*)
FROM users
WHERE username = 'testuser3';
| username | COUNT(*) |
|----------|----------|
| (null) | 0 |
Note that this query is not SQL standard compliant, and in strict mode you will get an error, because username is selected without beeng listed in the GROUP BY clause. (SQLFiddle).
The documentation says:
If you name columns to select in addition to the COUNT() value, a
GROUP BY clause should be present that names those same columns.
Otherwise, the following occurs:
If the ONLY_FULL_GROUP_BY SQL mode is enabled, an error occurs:
[...]
If ONLY_FULL_GROUP_BY is not enabled, the query is processed by
treating all rows as a single group, but the value selected for each
named column is indeterminate. The server is free to select the value
from any row:
[...]
Again: "The server is free to select the value from any row". But since there is no matching row, it returns NULL.

Answer 2

The fact that you receive "null" response is due to the LEFT JOIN statements. A Left join gets done, even if parameters do not match in that table, because they might make up columns in another (joined) table. To show only results that match criteria, use an INNER JOIN statement.
Also, you wish to COUNT(*) payments due by username, you should GROUP BY username I should think.
Try the following (untested, sorry, not in front of pc with editor on it)
SELECT u.username, u.first_name, u.email, COUNT(p.is_paid) as pending_payments
FROM users u
INNER JOIN students s ON u.id = s.user_id
INNER JOIN payments p ON s.user_id = p.student_id
WHERE u.username LIKE '%testuser%'
AND u.username IS NOT NULL
GROUP BY u.username
ORDER BY p.is_paid DESC, u.username ASC
Additionally, if you'd want to show only the results that have open pending payments you could add a HAVING statement after the GROUP BY, such as
HAVING u.pending_payments > 0
Tested with the following setup:
Below showing the query results. Second query does not have the HAVING statement.

Answer 3

I think the NULL values found for username etc. is mostly because of the missing group by in your queries.
I think it should work using:
SELECT users.username,
users.first_name,
users.email,
SUM(COALESCE( p.is_paid = 0, 0)) as pending_payments
FROM users
LEFT JOIN students s on s.user_id = users.id
LEFT JOIN payments p on p.student_id = s.id
WHERE users.username LIKE 'testuser3'
GROUP BY users.username,
users.first_name,
users.email;
And when leaving out the WHERE line, it will list for all the users:
SELECT users.username,
users.first_name,
users.email,
SUM(COALESCE( p.is_paid = 0, 0)) as pending_payments
FROM users
LEFT JOIN students s on s.user_id = users.id
LEFT JOIN payments p on p.student_id = s.id
GROUP BY users.username,
users.first_name,
users.email;
It will mention all users, even those who don't have any payment (pending or not) or users who are not listed in the Students table

Answer 4

You obviously has users with NULL username in users table. You can filter out the NULL value in the WHERE clause.
SELECT users.username, users.first_name, users.email,
COUNT(payments.is_paid = 0) as pending_payments
FROM users
LEFT JOIN students ON users.id = students.user_id
LEFT JOIN payments ON payments.student_id = students.id
WHERE users.username LIKE 'testuser' AND users.username IS NOT NULL;

Answer 5

Why cant use IS NOT NULL ?
SELECT users.username, users.first_name, users.email,
COUNT(payments.id) as pending_payments
FROM users
LEFT JOIN students ON users.id = students.user_id
LEFT JOIN payments ON payments.student_id = students.id AND payments.is_paid = 0
WHERE users.username LIKE 'testuser' and users.username is NOT NULL;