JPA 查询可以将结果作为 Java Map 返回吗? [英] Can a JPA Query return results as a Java Map?

问题描述

我们目前正在根据命名的 JPA 查询返回的两个字段手动构建一个 Map，因为 JPA 2.1 只提供了一个 getResultList() 方法:

We are currently building a Map manually based on the two fields that are returned by a named JPA query because JPA 2.1 only provides a getResultList() method:

@NamedQuery{name="myQuery",query="select c.name, c.number from Client c"}

HashMap<Long,String> myMap = new HashMap<Long,String>();

for(Client c: em.createNamedQuery("myQuery").getResultList() ){
     myMap.put(c.getNumber, c.getName);
}

但是，我觉得自定义映射器或类似的会更高效，因为此列表很容易包含 30,000 多个结果.

But, I feel like a custom mapper or similar would be more performant since this list could easily be 30,000+ results.

无需手动迭代即可构建地图的任何想法.

Any ideas to build a Map without iterating manually.

(我使用的是 OpenJPA，而不是休眠)

(I am using OpenJPA, not hibernate)

推荐答案

使用 JPA Query getResultStream 返回 Map 结果

从 JPA 2.2 版本开始，您可以使用 getResultStream Query 方法将 List 结果转换为 映射<整数，整数>:

Returning a Map result using JPA Query getResultStream

Since the JPA 2.2 version, you can use the getResultStream Query method to transform the List<Tuple> result into a Map<Integer, Integer>:

Map<Integer, Integer> postCountByYearMap = entityManager.createQuery("""
    select
       YEAR(p.createdOn) as year,
       count(p) as postCount
    from
       Post p
    group by
       YEAR(p.createdOn)
    """, Tuple.class)
.getResultStream()
.collect(
    Collectors.toMap(
        tuple -> ((Number) tuple.get("year")).intValue(),
        tuple -> ((Number) tuple.get("postCount")).intValue()
    )
);

使用 JPA 查询 getResultList 和 Java 流返回 Map 结果

如果您使用的是 JPA 2.1 或旧版本，但您的应用程序在 Java 8 或更新版本上运行，那么您可以使用 getResultList 并转换 List 到 Java 8 流:

Returning a Map result using JPA Query getResultList and Java stream

If you're using JPA 2.1 or older versions but your application is running on Java 8 or a newer version, then you can use getResultList and transform the List<Tuple> to a Java 8 stream:

Map<Integer, Integer> postCountByYearMap = entityManager.createQuery("""
    select
       YEAR(p.createdOn) as year,
       count(p) as postCount
    from
       Post p
    group by
       YEAR(p.createdOn)
    """, Tuple.class)
.getResultList()
.stream()
.collect(
    Collectors.toMap(
        tuple -> ((Number) tuple.get("year")).intValue(),
        tuple -> ((Number) tuple.get("postCount")).intValue()
    )
);

使用特定于 Hibernate 的 ResultTransformer 返回 Map 结果

另一种选择是使用 Hibernate Types 开源项目提供的 MapResultTransformer 类:

Map<Number, Number> postCountByYearMap = (Map<Number, Number>) entityManager.createQuery("""
    select
       YEAR(p.createdOn) as year,
       count(p) as postCount
    from
       Post p
    group by
       YEAR(p.createdOn)
    """)
.unwrap(org.hibernate.query.Query.class)
.setResultTransformer(
    new MapResultTransformer<Number, Number>()
)
.getSingleResult();

MapResultTransformer 适用于仍在运行的项目在 Java 6 上或使用较旧的 Hibernate 版本.

The MapResultTransformer is suitable for projects still running on Java 6 or using older Hibernate versions.

OP 说:

但是，我觉得自定义映射器或类似的会更高效因为这个列表很容易就有 30,000 多个结果.

But, I feel like a custom mapper or similar would be more performant since this list could easily be 30,000+ results.

这是一个糟糕的主意.您永远不需要选择 30k 条记录.这将如何适应 UI?或者，您为什么要对如此大量的记录进行操作?

This is a terrible idea. You never need to select 30k records. How would that fit in the UI? Or, why would you operate on such a large batch of records?

您应该使用查询分页，因为这将帮助您减少事务响应时间并提供更好的并发性.

You should use query pagination as this will help you reduce the transaction response time and provide better concurrency.

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