Удалите сгруппированный набор данных, когда общий счетчик равен нулю, с помощью подзапроса

Есть несколько способов сделать это, но чем менее подробен SQL, тем сложнее вам будет следовать логике. По этой причине я думаю, что использование нескольких общих табличных выражений позволит избежать необходимости использовать избыточные объединения, будучи наиболее удобочитаемыми.

-- assuming user_name and category_name are unique on [user] and [category] respectively.  

WITH valid_categories (category_id, category_name) AS 
(
    -- get set of valid categories
    SELECT c.category_id, c.category AS category_name
    FROM category c
    WHERE c.category_id NOT IN (19,20)
),
valid_users ([user_name]) AS 
(
    -- get set of users who belong to valid roles
    SELECT u.[user_name]
    FROM [user] u 
    WHERE EXISTS (
        SELECT *
        FROM [role] r
        WHERE u.id = r.[user_id] AND r.id IN (1,2)
    )

),
valid_entries (entry_id, [user_name], category_id, entry_count) AS
(
    -- provides a flag of 1 for easier aggregation
    SELECT e.[entry_id], e.[user_name], e.category_id, CAST( 1 AS INT) AS entry_count
    FROM [entry] e  
    WHERE e.entered_date BETWEEN TO_DATE('20140625','YYYYMMDD') AND TO_DATE('20140731', 'YYYYMMDD')
    -- determines if entry is within date range 
),
user_categories ([user_name], category_id, category_name) AS

(   SELECT u.[user_name], c.category_id, c.category_name
    FROM valid_users u
    -- get the cartesian product of users and categories
    CROSS JOIN valid_categories c
    -- get only users with a valid entry 
    WHERE EXISTS (
        SELECT *
        FROM valid_entries e
        WHERE e.[user_name] = u.[user_name]
    )
)

/*

You can use these for testing.

SELECT COUNT(*) AS valid_categories_count
FROM valid_categories

SELECT COUNT(*) AS valid_users_count
FROM valid_users

SELECT COUNT(*) AS valid_entries_count
FROM valid_entries

SELECT COUNT(*) AS users_with_entries_count
FROM valid_users u
WHERE EXISTS (
    SELECT *
    FROM user_categories uc
    WHERE uc.user_name = u.user_name
)

SELECT COUNT(*) AS users_without_entries_count
FROM valid_users u
WHERE NOT EXISTS (
    SELECT *
    FROM user_categories uc
    WHERE uc.user_name = u.user_name
)

SELECT uc.[user_name], uc.[category_name], e.[entry_count] 
FROM user_categories uc
INNER JOIN  valid_entries e ON (uc.[user_name] = e.[user_name] AND uc.[category_id] = e.[category_id])
*/

-- Finally, the results: 

SELECT uc.[user_name], uc.[category_name], SUM(NVL(e.[entry_count],0)) AS [entry_count]
FROM user_categories uc
LEFT OUTER JOIN  valid_entries e ON (uc.[user_name] = e.[user_name] AND uc.[category_id] = e.[category_id])