Ste*_*dge 46 java jpa distinct
DISTINCT在JPA中使用哪一列,是否可以更改它?
这是使用DISTINCT的示例JPA查询:
select DISTINCT c from Customer c
这没有多大意义 - 什么专栏是基于什么?它是否在实体上指定为注释,因为我找不到?
我想指定列来区分,例如:
select DISTINCT(c.name) c from Customer c
我正在使用MySQL和Hibernate.
Αλέ*_*κος 56
你很亲密
select DISTINCT(c.name) from Customer c
Tom*_*asz 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();
        }
}
kaz*_*aki 11
更新:请参阅最高投票的答案.
我自己现在已经过时了.因历史原因而留在这里.
在联接中通常需要HQL中的区别,而不是像您自己的简单示例中那样.
另请参见如何在HQL中创建不同的查询
Yan*_*ski 10
我同意kazanaki的回答,这对我很有帮助.我想选择整个实体,所以我用过
 select DISTINCT(c) from Customer c
在我的情况下,我有多对多的关系,我想在一个查询中加载具有集合的实体.
我使用了LEFT JOIN FETCH,最后我不得不将结果区分开来.
正如我在本文中所解释的,根据底层的JPQL或Criteria API查询类型,DISTINCT在JPA中具有两个含义。
对于返回标量投影的标量查询,例如以下查询:
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);
该DISTINCT关键字应传递给底层的SQL语句,因为我们希望之前,返回结果集数据库引擎过滤重复:
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]
对于实体查询,DISTINCT具有不同的含义。
如果不使用DISTINCT,则查询如下所示:
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())
);
将要加入post和这样的post_comment表:
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post记录在每个关联post_comment行的结果集中都是重复的。出于这个原因,List的Post实体将包含重复的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())
);
但是随后DISTINCT还传递给SQL查询,这是完全不希望的:
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Sort阶段,该阶段将增加开销而不会带来任何值,因为父子组合始终由于子PK列而返回唯一记录:
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
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]
执行计划将确认我们这次不再具有额外的排序阶段:
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
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
按照问题中的例子,它会是这样的。
SELECT DISTINCT new com.mypackage.MyNameType(c.name) from Customer c