Complex Query with related tables - Optimal Solution - mysql

Question

The Schema:
I have 3 Tables:
User
Feature
User_has_Feature:
initially all users has no features
Example data:
User:
| id | name |
| 1 | Rex |
| 2 | Job |
Feature:
| id | name |
| 1 | Eat |
| 2 | Walk |
User_has_Feature:
| id | user_id | feature_id | have_feature |
| 1 | 1 | 1 | true |
| 2 | 1 | 1 | true |
| 3 | 2 | 2 | true |
| 4 | 2 | 2 | false |
The questions are:
¿How to get only the records that have all features? (explicitly)
Example:
| user_name | feature_name | feature_status |
| Rex | Eat | true |
| Rex | Walk | true |
How to get records that do not have all the features? (again explicitly)
Example:
| user_name | feature_name | feature_status |
| Job | Eat | true |
| Job | Walk | false |
Some conditions have to be attended
I need the Users list with all features (true or false) in both queries like examples
User have 650k records (for now)
Feature have 45 records (for now)
Is one time query.
The idea is to export the result to a CSV file
Early Solution
thanks to the answers of (#RolandoMySQLDBA, #Tom Ingram, #DRapp) I found a solution:
SELECT u.name, f.name, IF(uhf.status=1,'YES','NO') as status
FROM user u
JOIN user_has_feature uhf ON u.id = uhf.user_id
JOIN feature f ON f.id = uhf.feature_id
JOIN
(
SELECT u.id as id
FROM user u
JOIN user_has_feature uhf ON uhf.user_id = u.id
WHERE uhf.status = 1
GROUP BY u.id
HAVING count(u.id) <= (SELECT COUNT(1) FROM feature)
) as `condition` ON `condition`.id = u.id
ORDER BY u.name, f.id, uhf.status
For get records that do not have all the features and for get all record that have all features change:
WHERE uhf.status = 1 by WHERE uhf.status = 2
HAVING count(u.id) <= (SELECT COUNT(1) FROM feature) by HAVING count(u.id) = (SELECT COUNT(1) FROM feature)
but I want to know if this is an optimal solution?

Answer 1

SELECT
UNF.*,
IF(
(LENGTH(UNF.FeatureList) - LENGTH(REPLACE(UNF.FeatureList,',','')))
= (FC.FeatureCount - 1),'Has All Features','Does Not Have All Features'
) HasAllFeatures
FROM
(SELECT
U.name user_name
GROUP_CONCAT(F.name) Features
FROM
(SELECT user_id,feature_id FROM User_has_Feature
WHERE feature_status = true) UHF
INNER JOIN User U ON UHF.user_id = U.id
INNER JOIN Feature F ON UHF.feature_id = F.id
GROUP BY
U.name
) UNF,
(SELECT COUNT(1) FeatureCount FROM Feature) FC
;
The UNF subquery returns with all users listed in User_has_Feature and a comma-separated list of the features. The column HasAllFeatures is determined by the number of columns in UNF.FeatureList. In your case, there are two features. If the number of commas in UNF.FeatureList is FeatureCount - 1, then the user has all features. Otherwise, user does not have all features.
Here is a better version that shows all users and whether or not they have all, some or no features
SELECT
U.name user_name,
IFNULL(UsersAndFeatures.HasAllFeatures,
'Does Not Have Any Features')
WhatFeaturesDoesThisUserHave
FROM
User U LEFT JOIN
(
SELECT
UHF.user_id id,
IF(
(LENGTH(UHF.FeatureList) - LENGTH(REPLACE(UHF.FeatureList,',','')))
= (FC.FeatureCount - 1),
'Has All Features',
'Does Not Have All Features'
) HasAllFeatures
FROM
(
SELECT user_id,GROUP_CONCAT(Feature.name) FeatureList
FROM User_has_Feature INNER JOIN Feature
ON User_has_Feature.feature_id = Feature.id
GROUP BY user_id
) UHF,
(SELECT COUNT(1) FeatureCount FROM Feature) FC
) UsersAndFeatures
USING (id);

Answer 2

select
u.id,
u.name as User_Name,
f.name as Feature_Name,
uhf.feature_Status
from
( select uhf.user_id,
sum( if( uhf.feature_status, 1, 0 ) ) as UserFeatureCount
from user_has_feature uhf
group by uhf.user_id ) AllUsersWithCounts
join
( select count(*) as AllFeaturesCount
from Feature ) AllFeatures
on AllUsersWithCounts.UserFeatureCount = AllFeatures.AllFeaturesCount
join user u
on AllUsersWithCounts.user_id = u.ID
join user_has_feature uhf
on AllUsersWithCounts.User_id = uhf.user_id
join feature f
on uhf.feature_id = f.id
The above query should get all people that explicitly have ALL features. In order to get those that do NOT have all features, just change the one join from = to <
on AllUsersWithCounts.UserFeatureCount < AllFeatures.AllFeaturesCount

Answer 3

Here's my bash at it
create a view of the general information
CREATE VIEW v_users_have_features AS
SELECT usr.id, usr.name, feature.name, has_feature.status
FROM usr
JOIN has_feature ON usr.id = has_feature.user_id
JOIN feature ON has_feature.feature_id = feature.id;
use the view for other queries
SELECT v_users_have_features.id, v_users_have_features.u_name, v_users_have_features.f_name
FROM v_users_have_features
GROUP BY v_users_have_features.id
HAVING COUNT( v_users_have_features.id ) = (SELECT COUNT( feature.id )
FROM feature
WHERE feature.name = v_users_have_features.f_name )
p.s. you may need to adapt (particularly the latter) to your exact requirements you could also omit creating the view and nest it in the FROM clause like in another answer it just seemed handier to create the view

Answer 4

Count the number of features. Write a query over users that uses a correlated subquery to find all the features a user has and count them. Make the restriction criterion in the top query the equality of that count and the global number of features.
Can MySQL do correlated subqueries? If not, you might need to use a better database.