JPA 2.0：加载实体的字段子集 [英] JPA 2.0: Load a subset of fields for an entity

问题描述

我有一个实体名称地址。地址实体有几个字段，其中之一是CITY。我通过调用entityManager.createQuery创建了一个查询，但我的查询仅包含select子句中的CITY字段（因为我只想加载该字段）。 CITY字段的类型是String。当我得到我的resultList时，我没有得到一个Address对象列表，而是一个Object []列表。

是否有任何方法可以创建地址列表而不是Object []的列表？我的JPA提供程序是休眠，最新版本。我想要一个解决方案，不需要使用Hibernate特有的任何东西。

解决方案

在查询中，这可以通过构造函数表达式实现。通常情况下，您可以使用自定义的DTO，但它也应该与实体一起工作。首先在你的实体中创建一个额外的构造函数，只需要所需的字段：

  public Address（）{
 // default构造函数
} 
 
 public地址（String city）{
 this.city = city; 
 
 $ / code> 

您的查询可能如下所示：

 从地址a中选择新的your.package.Address（a.city）... 
  

请注意，生成的对象不会附加到EntityManager，所以对它的更改不会自动保持。

I have an Entity named address. The address Entity has several fields, one of which is CITY. I've created a query by calling entityManager.createQuery but my query only includes the CITY field in the select clause (because I only want to load that field). The CITY field is of type String. When I get my resultList, I do not get a list of Address objects but instead a list of Object[].

Is there any way to have a list of Address created instead of a list of Object[]? My JPA provider is hibernate, latest version. I want a solution that does not require the use of anything Hibernate specific.

解决方案

This is possible with constructor expressions in your query. Normally you would use this with a custom DTO, but it should work with the entity too. First create an additional constructor in your entity taking only the needed fields:

public Address() {
    //default constructor
}

public Address(String city) {
    this.city = city;
}

Your query could then look like this:

select new your.package.Address(a.city) from Address a where ...

Note that the resulting object is not attached to the EntityManager, so changes to it won't be automatically persisted.

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