使用Query DSL从每个组中选择具有最大值的记录 [英] Select record with max value from each group with Query DSL

问题描述

我有一个得分表，其中有球员得分，我想为每个得分最高的球员选择唯一的记录.

I have a score table where I have players scores, and I want select unique records for each player with the biggest score.

这是桌子:

id | player_id | score | ...
1  | 1         |  10   | ...
2  | 2         |  21   | ...
3  | 3         |  9    | ...
4  | 1         |  30   | ...
5  | 3         |  2    | ...

预期结果:

id | player_id | score | ...
2  | 2         |  21   | ...
3  | 3         |  9    | ...
4  | 1         |  30   | ...

我可以这样用纯SQL来实现:

I can achieve that with pure SQL like this:

SELECT *
FROM player_score ps
WHERE ps.score = 
(
    SELECT max(ps2.score)
    FROM player_score ps2
    WHERE ps2.player_id = ps.player_id
) 

您能告诉我如何使用查询dsl实现相同的查询吗?我找到了一些使用JPASubQuery的解决方案，但该类对我不起作用(我的IDE无法解析该类).我正在使用querydsl4.x.预先谢谢你.

Can you tell me how to achieve the same query with query dsl? I found some solutions with JPASubQuery but this class doesn't work for me (my IDE cannot resolve this class). I am using querydsl 4.x. Thank you in advance.

推荐答案

JPASubQuery已在querydsl 4中删除.请改为使用JPAExpressions.select.您的WHERE子句应如下所示:

JPASubQuery has been removed in querydsl 4. Instead use JPAExpressions.select. Your WHERE clause should look something like this:

.where(playerScore.score.eq(JPAExpressions.select(playerScore2.score.max())
                            .from(playerScore2))
                            .where(playerScore2.playerId.eq(playerScore.playerId)))

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