62

DISTINCT 在 JPA 中使用什么列，是否可以更改它？

这是使用 DISTINCT 的示例 JPA 查询：

select DISTINCT c from Customer c

哪个没有多大意义——不同的列基于什么？它是否在实体上指定为注释，因为我找不到？

我想指定要区分的列，例如：

select DISTINCT(c.name) c from Customer c

我正在使用 MySQL 和 Hibernate。

4

6 回答 6

72

你很亲密。

select DISTINCT(c.name) from Customer c
于 2012-10-24T13:58:10.930 回答
23

根据底层的 JPQL 或 Criteria API 查询类型，DISTINCT在 JPA 中有两个含义。

标量查询

对于返回标量投影的标量查询，如以下查询：

List<Integer> publicationYears = entityManager
.createQuery(
    "select distinct year(p.createdOn) " +
    "from Post p " +
    "order by year(p.createdOn)", Integer.class)
.getResultList();

LOGGER.info("Publication years: {}", publicationYears);

应该将DISTINCT关键字传递给底层 SQL 语句，因为我们希望数据库引擎在返回结果集之前过滤重复项：

SELECT DISTINCT
    extract(YEAR FROM p.created_on) AS col_0_0_
FROM
    post p
ORDER BY
    extract(YEAR FROM p.created_on)

-- Publication years: [2016, 2018]

实体查询

对于实体查询，DISTINCT有不同的含义。

不使用DISTINCT，查询如下：

List<Post> posts = entityManager
.createQuery(
    "select p " +
    "from Post p " +
    "left join fetch p.comments " +
    "where p.title = :title", Post.class)
.setParameter(
    "title", 
    "High-Performance Java Persistence eBook has been released!"
)
.getResultList();

LOGGER.info(
    "Fetched the following Post entity identifiers: {}", 
    posts.stream().map(Post::getId).collect(Collectors.toList())
);

将加入这样post的post_comment表格：

SELECT p.id AS id1_0_0_,
       pc.id AS id1_1_1_,
       p.created_on AS created_2_0_0_,
       p.title AS title3_0_0_,
       pc.post_id AS post_id3_1_1_,
       pc.review AS review2_1_1_,
       pc.post_id AS post_id3_1_0__
FROM   post p
LEFT OUTER JOIN
       post_comment pc ON p.id=pc.post_id
WHERE
       p.title='High-Performance Java Persistence eBook has been released!'

-- Fetched the following Post entity identifiers: [1, 1]

但是父记录在每个关联行post的结果集中重复。post_comment因此，List实体的Post将包含重复的Post实体引用。

为了消除Post实体引用，我们需要使用DISTINCT：

List<Post> posts = entityManager
.createQuery(
    "select distinct p " +
    "from Post p " +
    "left join fetch p.comments " +
    "where p.title = :title", Post.class)
.setParameter(
    "title", 
    "High-Performance Java Persistence eBook has been released!"
)
.getResultList();
 
LOGGER.info(
    "Fetched the following Post entity identifiers: {}", 
    posts.stream().map(Post::getId).collect(Collectors.toList())
);

但是 thenDISTINCT也被传递给 SQL 查询，这根本不可取：

SELECT DISTINCT
       p.id AS id1_0_0_,
       pc.id AS id1_1_1_,
       p.created_on AS created_2_0_0_,
       p.title AS title3_0_0_,
       pc.post_id AS post_id3_1_1_,
       pc.review AS review2_1_1_,
       pc.post_id AS post_id3_1_0__
FROM   post p
LEFT OUTER JOIN
       post_comment pc ON p.id=pc.post_id
WHERE
       p.title='High-Performance Java Persistence eBook has been released!'
 
-- Fetched the following Post entity identifiers: [1]

通过传递DISTINCT给 SQL 查询，EXECUTION PLAN 将执行一个额外的排序阶段，这会增加开销而不会带来任何价值，因为由于子 PK 列，父子组合总是返回唯一记录：

Unique  (cost=23.71..23.72 rows=1 width=1068) (actual time=0.131..0.132 rows=2 loops=1)
  ->  Sort  (cost=23.71..23.71 rows=1 width=1068) (actual time=0.131..0.131 rows=2 loops=1)
        Sort Key: p.id, pc.id, p.created_on, pc.post_id, pc.review
        Sort Method: quicksort  Memory: 25kB
        ->  Hash Right Join  (cost=11.76..23.70 rows=1 width=1068) (actual time=0.054..0.058 rows=2 loops=1)
              Hash Cond: (pc.post_id = p.id)
              ->  Seq Scan on post_comment pc  (cost=0.00..11.40 rows=140 width=532) (actual time=0.010..0.010 rows=2 loops=1)
              ->  Hash  (cost=11.75..11.75 rows=1 width=528) (actual time=0.027..0.027 rows=1 loops=1)
                    Buckets: 1024  Batches: 1  Memory Usage: 9kB
                    ->  Seq Scan on post p  (cost=0.00..11.75 rows=1 width=528) (actual time=0.017..0.018 rows=1 loops=1)
                          Filter: ((title)::text = 'High-Performance Java Persistence eBook has been released!'::text)
                          Rows Removed by Filter: 3
Planning time: 0.227 ms
Execution time: 0.179 ms

带有 HINT_PASS_DISTINCT_THROUGH 的实体查询

为了从执行计划中消除排序阶段，我们需要使用HINT_PASS_DISTINCT_THROUGHJPA 查询提示：

List<Post> posts = entityManager
.createQuery(
    "select distinct p " +
    "from Post p " +
    "left join fetch p.comments " +
    "where p.title = :title", Post.class)
.setParameter(
    "title", 
    "High-Performance Java Persistence eBook has been released!"
)
.setHint(QueryHints.HINT_PASS_DISTINCT_THROUGH, false)
.getResultList();
 
LOGGER.info(
    "Fetched the following Post entity identifiers: {}", 
    posts.stream().map(Post::getId).collect(Collectors.toList())
);

现在，SQL 查询将不包含DISTINCT但Post实体引用重复项将被删除：

SELECT
       p.id AS id1_0_0_,
       pc.id AS id1_1_1_,
       p.created_on AS created_2_0_0_,
       p.title AS title3_0_0_,
       pc.post_id AS post_id3_1_1_,
       pc.review AS review2_1_1_,
       pc.post_id AS post_id3_1_0__
FROM   post p
LEFT OUTER JOIN
       post_comment pc ON p.id=pc.post_id
WHERE
       p.title='High-Performance Java Persistence eBook has been released!'
 
-- Fetched the following Post entity identifiers: [1]

执行计划将确认这次我们不再有额外的排序阶段：

Hash Right Join  (cost=11.76..23.70 rows=1 width=1068) (actual time=0.066..0.069 rows=2 loops=1)
  Hash Cond: (pc.post_id = p.id)
  ->  Seq Scan on post_comment pc  (cost=0.00..11.40 rows=140 width=532) (actual time=0.011..0.011 rows=2 loops=1)
  ->  Hash  (cost=11.75..11.75 rows=1 width=528) (actual time=0.041..0.041 rows=1 loops=1)
        Buckets: 1024  Batches: 1  Memory Usage: 9kB
        ->  Seq Scan on post p  (cost=0.00..11.75 rows=1 width=528) (actual time=0.036..0.037 rows=1 loops=1)
              Filter: ((title)::text = 'High-Performance Java Persistence eBook has been released!'::text)
              Rows Removed by Filter: 3
Planning time: 1.184 ms
Execution time: 0.160 ms
于 2018-11-21T06:03:04.903 回答
13 
@Entity
@NamedQuery(name = "Customer.listUniqueNames", 
            query = "SELECT DISTINCT c.name FROM Customer c")
public class Customer {
        ...

        private String name;

        public static List<String> listUniqueNames() {
             return = getEntityManager().createNamedQuery(
                   "Customer.listUniqueNames", String.class)
                   .getResultList();
        }
}
于 2012-08-28T22:44:58.687 回答
12

更新：请查看投票最多的答案。

我自己的现在已经过时了。仅出于历史原因保留在这里。

连接中通常需要 HQL 中的不同，而不是像您自己这样的简单示例。

另请参阅如何在 HQL 中创建 Distinct 查询

于 2009-08-28T15:40:50.737 回答
11

我同意kazanaki的回答，它帮助了我。我想选择整个实体，所以我用

 select DISTINCT(c) from Customer c

就我而言，我有多对多的关系，我想在一个查询中加载带有集合的实体。

我使用了 LEFT JOIN FETCH，最后我必须使结果与众不同。

于 2016-10-19T15:45:14.473 回答
4

我会使用 JPA 的构造函数表达式功能。另请参阅以下答案：

JPQL 构造函数表达式 - org.hibernate.hql.ast.QuerySyntaxException：表未映射

按照问题中的示例，它将是这样的。

SELECT DISTINCT new com.mypackage.MyNameType(c.name) from Customer c
于 2017-06-29T06:32:04.223 回答