Count for all values of enum in PostgreSQL

Question

I have a table called users , which has the following columns:

id: INT NOT NULL
face: face_type

face_type is an ENUM type that has the following values: 'square' , 'round' and 'triangle' .

And I have another table called houses , which has the following columns:

id: INT NOT NULL
user_id: INT NOT NULL

Now, I want to get all the houses grouped by the different type of face types. So, what I have so far is this:

SELECT users.face_type, COUNT(*)
FROM users
LEFT JOIN houses ON houses.user_id = users.id
GROUP BY users.face_type

The problem is that I also want to get rows for face_type which none of the users have, as well as a result for NULL face_type. So, for example, if I have the following data:

users (id, face_type)

1, 'round'
2, 'triangle'

houses (id, user_id)
1, 1
2, 1
3, 2

I would expect the result to be:

face_type, count
'round'          2
'triangle'       1
'square'         0
null             0

I know how to get all the potential values of the face_type ENUM, by doing :

SELECT unnest(enum_range(NULL::face_type)) AS face_types;

But I don't know how to use that to count all potential face types in the aggregate, as well as also calculating for NULL face types.

Answer 1

You can use LEFT JOIN :

SELECT ft.face_type, COUNT(h.user_id)
FROM (SELECT unnest(enum_range(NULL::face_type)) AS face_types
     ) ft LEFT JOIN
     users u
     ON u.face_type = ft.face_type LEFT JOIN
     houses h
     ON h.user_id = u.id
GROUP BY ft.face_type;

To get NULL , just use UNION ALL :

SELECT ft.face_type, COUNT(h.user_id)
FROM (SELECT unnest(enum_range(NULL::face_type)) AS face_types
      UNION ALL
      SELECT NULL
     ) ft LEFT JOIN
     users u
     ON u.face_type = ft.face_type LEFT JOIN
     houses h
     ON h.user_id = u.id
GROUP BY ft.face_type;

Of course, the = will not every match. If that is possible, then you want to change the JOIN condition to u.face_type is not distinct from ft.face_type .

Answer 2

to COUNT(houses.*)

SELECT face_type.type, COUNT(houses.*)
FROM (SELECT unnest(enum_range(NULL::face_type))) AS face_type(type)
FULL JOIN users ON users.face_type=face_type.type
LEFT JOIN houses ON houses.user_id = users.id
GROUP BY face_type.type

Answer 3

A LEFT JOIN starting from the ENUM and going to users and houses will allow you to recover totals for each enumerated value. To also display the NULL face types, you can use a UNION query.

SELECT 
    ft.face_type,
    COUNT(ho.user_id) as cnt
FROM 
    (SELECT unnest(enum_range(NULL::face_type)) AS face_types) ft
    LEFT JOIN users us ON us.face_type = ft.fact_type
    LEFT JOIN houses ho ON ho.user_id = us.id
GROUP BY ft.face_type
UNION
SELECT
    null,
    COUNT(ho.user_id)
FROM houses ho
INNER JOIN users us ON ho.user_id = us.id AND us.face_type IS NULL
ORDER BY cnt desc