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nahjjun
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Jan 23, 2026
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nahjjun
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nahjjun
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전체적으로 N+1 문제만 해결하면 될 것 같습니다!! 코드 수정하느라 고생하셨습니다!!!
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| .findBySenderAndStatus(sender, FriendRequestStatus.PENDING) | ||
| .stream() | ||
| .map(req -> FriendRequestListResDto.builder() | ||
| .requestId(req.getId()) | ||
| .userId(req.getReceiver().getId()) | ||
| .nickname(req.getReceiver().getNickname()) | ||
| .requestedAt(req.getCreatedAt()) | ||
| .build() | ||
| ) | ||
| .map(req -> { | ||
| User receiver = req.getReceiver(); |
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FriendRequestRepository의 findBySenderAndStatus() 함수에서 JPQL로 Fetch Join으로 Receiver도 한번에 가져오는 것이 좋을 것 같습니다.
현재 구조에서는 매 FriendRequest에 연결되어있는 Receiver를 Lazy Loading으로 매번 쿼리문을 가져오기 때문에 n+1 문제가 우려됩니다. 해당 부분 수정해주시면 성능적으로 이점이 있을 것 같습니다!
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| List<FriendResDto> friends = friendRepository.findByUser(user) | ||
| .stream() | ||
| .map(friend -> FriendResDto.builder() | ||
| .friendId(friend.getFriend().getId()) | ||
| .nickname(friend.getFriend().getNickname()) | ||
| .build() | ||
| ) | ||
| .map(friend -> { | ||
| var friendUser = friend.getFriend(); |
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이 부분도 위에서 언급한 것과 마찬가지로, findByUser 함수에서 Friend 엔티티도 Fetch Join으로 가져와야 N+1 문제가 생기지 않을 것 같습니다!
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