JPA CriteriaQuery Join-如何在子元素上进行联接? [英] JPA CriteriaQuery Join - how to join on a subelement?

问题描述

我正在尝试使用条件查询来执行联接.

I am trying to use criteria queries to perform a join.

我的班级结构是这样的:

My class structure is like this:

@Entity
class Parent {
  Intermediate intermediate;
}

@Entity
class Intermediate {
  Set<Child> children
}

@Entity
class Child {
  String someProperty;
}

我正在尝试让所有父母中至少一个孩子具有匹配的财产，但无法确定如何加入.如果没有中间对象，那将很容易:

I am trying to get all parents where at least one child has a matching property, but cannot work out how to do the joining. If the Intermediate object wasn't there, it would be quite easy:

// Here, Intermediate doesn't exist - Parent has the Set<Child> property

CriteriaBuilder builder = ...
CriteriaQuery<Parent> query = ...
Root<Parent> root = query.from( Parent.class );
Join<Parent, Child> join = root.join( "child" );

Path path = join.get( "someProperty" );
Predicate predicate = builder.equal( path, "somevalue" );

但是使用中间实体这样做会破坏它

But doing this with the intermediate entity breaks it

CriteriaBuilder builder = ...
CriteriaQuery<Parent> query = ...
Root<Parent> root = query.from( Parent.class );
Join<Parent, Child> join = root.join( "intermediate.child" ); <-- Fails

Caused by: java.lang.IllegalArgumentException: Unable to resolve attribute [intermediate.child] against path
    at org.hibernate.ejb.criteria.path.AbstractPathImpl.unknownAttribute(AbstractPathImpl.java:120) ~[hibernate-entitymanager-4.2.0.Final.jar:4.2.0.Final]
    at org.hibernate.ejb.criteria.path.AbstractPathImpl.locateAttribute(AbstractPathImpl.java:229) ~[hibernate-entitymanager-4.2.0.Final.jar:4.2.0.Final]
    at org.hibernate.ejb.criteria.path.AbstractFromImpl.join(AbstractFromImpl.java:411) ~[hiberna

te-entitymanager-4.2.0.Final.jar:4.2.0.Final]

te-entitymanager-4.2.0.Final.jar:4.2.0.Final]

在此示例中，我可以为孩子设计Path对象，但是JPA似乎不想让我使用Path对象而不是Root对象进行Join.

I can work out the Path object for the child in this example, but JPA doesn't seem to want to let me make a Join using a Path object instead of a Root one.

有人可以帮我吗?谢谢

推荐答案

我认为您的课程必须是这样的.

I think your classes need to be something like this.

@Entity
class Parent {
  @OneToOne
  Intermediate intermediate;
}

@Entity
class Intermediate {
  @OneToOne(mappedBy="intermediate")
  Parent parent;
  @OneToOne
  Set<Child> children
}

@Entity
class Child {
  String someProperty;
}

然后您可以按照以下步骤进行连接.

Then you can do the join as it follows.

CriteriaBuilder builder = ...
CriteriaQuery<Parent> query = ...
Root<Parent> root = query.from( Parent.class );
Join<Parent, Child> join = root.join("intermediate").join("children");

这篇关于JPA CriteriaQuery Join-如何在子元素上进行联接?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！