How do I make a conditional join? - postgresql

Question

I have the following query
SELECT p.first_name || ' ' || p.last_name as full_name, p.user_id,
u.email, u.username,
(
SELECT EXTRACT (year from (age(now(), p.date_of_birth)))
)age,
co.name as country, s.name as state, c.name as city,
(
SELECT country.phonecode FROM countries country WHERE ph.country_id = country.id
)phone_code, ph.area_code, ph.number as phone_number,
(
SELECT i.path FROM images i WHERE i.id = p.image_id
)avatar
FROM people p
JOIN users u ON u.id = p.user_id
JOIN addresses a ON a.id = p.address_id
JOIN cities c ON c.id = a.city_id
JOIN states s ON s.id = c.state_id
JOIN countries co ON co.id = s.country_id
JOIN phones ph ON ph.person_id = p.id
WHERE (
SELECT u.status FROM users u WHERE u.id = p.user_id
) = 'U'
ORDER BY p.id DESC
However, some people don't necessarily have information in Address and Phone and I want to bring that information anyway, how could I do it?
Little example:
Users
+----+-----------+-----------+---------------+-------------------------+
| id | username | role | status | email |
+----+-----------+-----------+---------------+-------------------------+
| 7 | Shibari | P | U | deathwrap#live.ca |
| 12 | Josh23 | P | U | heeman#gmail.ca |
| 13 | Test22 | P | U | test#gmail.ca |
+----+-----------+-----------+---------------+-------------------------+
People
+----+----------+-----------+-------------+-------+------------+----------+
| id |first_name| last_name |date_of_birth|user_id| address_id | image_id |
+----+----------+-----------+-------------+-------+------------+----------+
| 7 | Edward | Test | 1983-10-13 | 7 | 11 | [null] |
| 12 | Josh | Test | 1996-04-11 | 12 | [null] | 7 |
| 13 | Gavin | Cringe | 1991-04-11 | 13 | 18 | 8 |
+----+----------+-----------+-------------+-------+-----------------------+
Addresses
+----+---------------------+-----------+---------------+
| id | name | zip_code | city_id |
+----+---------------------+-----------+---------------+
| 11 | Rue De La Fonderie |[null] | 26497 |
| 18 | Av. Test |[null] | 6486 |
+----+---------------------+-----------+---------------+
Cities
+-----+---------------------+-----------+
| id | name | state_id |
+-----+---------------------+-----------+
|6486 | Resistencia |210 |
|26497| Luxembourg |2237 |
+-----+---------------------+-----------+
States
+-----+---------------------+-----------+
| id | name |country_id |
+-----+---------------------+-----------+
|210 | Chaco |10 |
|2237 | Luxembourg |127 |
+-----+---------------------+-----------+
Countries
+-----+---------------------+-----------+
| id | name | phonecode |
+-----+---------------------+-----------+
|10 | Argentina |54 |
|127 | Luxembourg |352 |
+-----+---------------------+-----------+
Phones
+----+-----------+-----------+---------------+-------------+
| id | area_code | number | person_id | country_id |
+----+-----------+-----------+---------------+-------------+
| 6 | 661 | 12345 | 7 | 127 |
+----+-----------+-----------+---------------+-------------+
Images
+-----+----------------+
| id | path |
+-----+----------------+
|7 | 7.jpg |
|8 | 8.jpg |
+-----+----------------+
Here's the output I desire
+------------+-------+-----------------+--------+---+----------+----------+-----------+----------+
| full_name |user_id| email |username|age|country |state |city |phone_code|
+------------+-------+-----------------+--------+---+----------+----------+-----------+----------+
|Edward Test |7 |deathwrap#live.ca|Shibari |37 |Luxembourg|Luxembourg|Luxembourg |352 |
|Josh Test |12 |heeman#gmail.ca |Josh23 |24 |[null] |[null] |[null] |[null] |
|Gavin Cringe|13 |test#gmail.ca |Test22 |29 |Argentina |Chaco |Resistencia|[null] |
+------------+-------+-----------------+--------+---+----------+----------+-----------+----------+
+------------+------------+-----------------+
| area_code |phone_number| avatar |
+------------+------------+-----------------+
|661 |12345 | [null] |
|[null] |[null] | 7.jpg |
|[null] |[null] | 8.jpg |
+------------+------------+-----------------+
At the moment it's just printing the first entry because it has both an address and a phone number linked, but I want to print this info even when there's no phone number or address linked, how could I do this?

Answer 1

Use left joins instead of inner joins:
SELECT ...
FROM people p
LEFT JOIN users u ON u.id = p.user_id AND u.status = 'U'
LEFT JOIN addresses a ON a.id = p.address_id
LEFT JOIN cities c ON c.id = a.city_id
LEFT JOIN states s ON s.id = c.state_id
LEFT JOIN countries co ON co.id = s.country_id
LEFT JOIN phones ph ON ph.person_id = p.id
Using the above left join approach should ensure that every person record appear in the output even if it doesn't match to one or more of the otber tables. Note that I have moved the check on status from the WHERE clause to the ON clause of the left join with the users table.