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[Fix] 내가 다녀온 산 목록 API에서 imageUrl null 반환 수정 #159

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Merged
howooyeon merged 1 commit into from
May 29, 2026
Merged

[Fix] 내가 다녀온 산 목록 API에서 imageUrl null 반환 수정 #159

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4 changes: 2 additions & 2 deletions src/main/java/com/semosan/api/domain/hiking/repository/HikingRecordRepository.java
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Original file line number Diff line number Diff line change
Expand Up @@ -35,14 +35,14 @@ AND NOT EXISTS (
SELECT
hr.mountain_id AS mountainId,
m.name AS mountainName,
m.image_url AS imageUrl,
m.image_urls->>0 AS imageUrl,
COUNT(hr.id) AS hikingCount,
MAX(hr.created_at) AS lastHikedAt
FROM hiking_records hr
JOIN hiking_members hm ON hm.hiking_record_id = hr.id
JOIN mountains m ON m.id = hr.mountain_id
WHERE hm.user_id = :userId
GROUP BY hr.mountain_id, m.name, m.image_url
GROUP BY hr.mountain_id, m.name, m.image_urls->>0
ORDER BY MAX(hr.created_at) DESC, hr.mountain_id DESC
Comment on lines 35 to 46

@gemini-code-assist gemini-code-assist Bot May 29, 2026

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medium

PostgreSQL에서는 GROUP BY 절에 테이블의 기본 키(Primary Key)인 m.id를 지정하면, 해당 테이블의 다른 컬럼들(m.name, m.image_urls->>0 등)은 함수적 종속성(Functional Dependency) 덕분에 GROUP BY 절에 개별적으로 나열하지 않고도 SELECT 절에서 자유롭게 조회할 수 있습니다.\n\n현재 쿼리에서는 m.image_urls->>0과 같은 복잡한 표현식이 SELECT와 GROUP BY 양쪽에 중복되어 작성되어 있어, 향후 이미지 조회 로직이 변경될 때 두 곳 모두 수정해야 하는 유지보수 상의 번거로움이 있습니다.\n\n따라서 SELECT 절에서 hr.mountain_id 대신 m.id를 조회하고, GROUP BY 절을 m.id로 단순화하면 쿼리의 가독성과 유지보수성을 크게 향상시킬 수 있습니다.

Suggested change
SELECT
hr.mountain_id AS mountainId,
m.name AS mountainName,
m.image_url AS imageUrl,
m.image_urls->>0 AS imageUrl,
COUNT(hr.id) AS hikingCount,
MAX(hr.created_at) AS lastHikedAt
FROM hiking_records hr
JOIN hiking_members hm ON hm.hiking_record_id = hr.id
JOIN mountains m ON m.id = hr.mountain_id
WHERE hm.user_id = :userId
GROUP BY hr.mountain_id, m.name, m.image_url
GROUP BY hr.mountain_id, m.name, m.image_urls->>0
ORDER BY MAX(hr.created_at) DESC, hr.mountain_id DESC
SELECT\n m.id AS mountainId,\n m.name AS mountainName,\n m.image_urls->>0 AS imageUrl,\n COUNT(hr.id) AS hikingCount,\n MAX(hr.created_at) AS lastHikedAt\n FROM hiking_records hr\n JOIN hiking_members hm ON hm.hiking_record_id = hr.id\n JOIN mountains m ON m.id = hr.mountain_id\n WHERE hm.user_id = :userId\n GROUP BY m.id\n ORDER BY MAX(hr.created_at) DESC, m.id DESC
References
  1. 우선순위 가이드에 따라 유지보수성(Maintainability) 및 가독성(Readability)을 개선하기 위해 중복된 표현식을 제거하고 쿼리를 단순화합니다. (link)

""",
countQuery = """
Expand Down