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);