# 모든 범주에 연결된 모든 그룹

**URL:** https://meta.discourse.org/t/all-groups-associated-with-all-categories/282604
**Category:** Data & reporting
**Tags:** groups, sql-query
**Created:** [10월 18, 2023, 7:09오전 UTC](https://meta.discourse.org/t/all-groups-associated-with-all-categories/282604 "2023-10-18T07:09:26Z")
**Posts on this page:** 2
**Page:** 1

<div class="post-metadata">

### Author: ![Srinivas\_Chilukuri1](https://sea3.discourse-cdn.com/meta/user_avatar/meta.discourse.org/srinivas_chilukuri1/32/247096_2.png) [@Srinivas\_Chilukuri1](https://meta.discourse.org/u/Srinivas_Chilukuri1)
#### Post date: [10월 18, 2023, 7:09오전 UTC](https://meta.discourse.org/t/all-groups-associated-with-all-categories/282604/1 "2023-10-18T07:09:26Z")

</div>

아래 쿼리에 대한 도움이 필요합니다.

- 포럼의 모든 카테고리에서, 해당 카테고리에 접근 권한이 있는 모든 그룹을 카테고리별로 나열

---

<div class="post-metadata">

### Author: ![JammyDodger](https://sea3.discourse-cdn.com/meta/user_avatar/meta.discourse.org/jammydodger/32/254611_2.png) [@JammyDodger](https://meta.discourse.org/u/JammyDodger)
#### Post date: [10월 18, 2023, 3:25오후 UTC](https://meta.discourse.org/t/all-groups-associated-with-all-categories/282604/2 "2023-10-18T15:25:37Z")

</div>

이런 식으로 하면 될까요:

```sql
SELECT 
    cg.category_id,
    g.name AS "Group Name",
    CASE 
      WHEN cg.permission_type = 1 THEN 'Create'
      WHEN cg.permission_type = 2 THEN 'Reply'
      WHEN cg.permission_type = 3 THEN 'See'
    END AS "Permission"
FROM category_groups cg
  JOIN groups g ON g.id = cg.group_id
ORDER BY cg.category_id

```