728x90
SELECT
*
# info.ITEM_ID,
# info.ITEM_NAME,
# info.RARITY
FROM ITEM_INFO AS info
# JOIN ITEM_TREE AS tree
# ON info.ITEM_ID = tree.ITEM_ID
# JOIN ITEM_INFO AS parent
# ON tree.PARENT_ITEM_ID = parent.ITEM_ID
# WHERE parent.RARITY = 'RARE'
# ORDER BY info.ITEM_ID DESC
SELECT
*
# info.ITEM_ID,
# info.ITEM_NAME,
# info.RARITY
FROM ITEM_INFO AS info
JOIN ITEM_TREE AS tree
ON info.ITEM_ID = tree.ITEM_ID
# JOIN ITEM_INFO AS parent
# ON tree.PARENT_ITEM_ID = parent.ITEM_ID
# WHERE parent.RARITY = 'RARE'
# ORDER BY info.ITEM_ID DESC
SELECT
*
# info.ITEM_ID,
# info.ITEM_NAME,
# info.RARITY
FROM ITEM_INFO AS info
JOIN ITEM_TREE AS tree
ON info.ITEM_ID = tree.ITEM_ID
JOIN ITEM_INFO AS parent
ON tree.PARENT_ITEM_ID = parent.ITEM_ID
# WHERE parent.RARITY = 'RARE'
# ORDER BY info.ITEM_ID DESC
- 여기서 tree.PARENT_ITEM_ID와 parent.ITEM_ID는 컬럼명은 다르지만, 둘의 값이 같을 때 매칭돼서 조인이 이루어진다
- PARENT_ITEM_ID에 들어 있는 값은 실제 부모 아이템의 ID이고, 그걸 ITEM_INFO에서 찾는 것
SELECT
*
# info.ITEM_ID,
# info.ITEM_NAME,
# info.RARITY
FROM ITEM_INFO AS info
JOIN ITEM_TREE AS tree
ON info.ITEM_ID = tree.ITEM_ID
JOIN ITEM_INFO AS parent
ON tree.PARENT_ITEM_ID = parent.ITEM_ID
WHERE parent.RARITY = 'RARE'
# ORDER BY info.ITEM_ID DESC
SELECT
info.ITEM_ID,
info.ITEM_NAME,
info.RARITY
FROM ITEM_INFO AS info
JOIN ITEM_TREE AS tree
ON info.ITEM_ID = tree.ITEM_ID
JOIN ITEM_INFO AS parent
ON tree.PARENT_ITEM_ID = parent.ITEM_ID
WHERE parent.RARITY = 'RARE'
ORDER BY info.ITEM_ID DESC
난는...SQL이 너무 어렵다...
728x90
'Programming > SQL' 카테고리의 다른 글
[LeetCode] 175. Combine Two Tables (0) | 2025.05.05 |
---|---|
[프로그래머스] 이름에 el이 들어가는 동물 찾기 (0) | 2025.04.13 |
[프로그래머스] 최솟값 구하기 (0) | 2025.04.12 |
[프로그래머스] 가격이 제일 비싼 식품의 정보 출력하기 (0) | 2025.04.10 |
[프로그래머스] 가격대 별 상품 개수 구하기 (0) | 2025.04.09 |