Java 使用单表继承对实体层次结构进行JPA条件查询

假设我拥有以下实体：

@Entity
@Inheritance(strategy = SINGLE_TABLE)
@DiscriminatorColumn(name = "type")
public abstract class BaseEntity { 
    private Date someDate;
    private Date otherDate;
    private boolean flag;
}

@Entity
@DiscriminatorValue("entity1")
public class Entity1 extends BaseEntity { 
    private String someProperty;
}

@Entity
@DiscriminatorValue("entity2")
public class Entity2 extends BaseEntity { 
    private String otherProperty;
}

我正在尝试构建一个条件查询，该查询根据BaseEntity和这两个子类中的属性返回BaseEntity的实例。所以本质上我在寻找一个与这个伪SQL对应的条件查询：

SELECT * FROM <BaseEntity table name>
WHERE someDate < ? AND otherDate > ? AND flag = ?
AND someProperty = ? AND otherProperty = ?;

从中选择*
什么地方？其他日期>？和flag=？
还有someProperty=？和其他属性=？；

我不想构建两个单独的查询，因为它们有太多的重叠（即大多数属性都在基类中）。但是，如果我将BaseEntity声明为根，我还没有找到在查询中引用子类属性的方法。有可能构建这样的条件查询吗

更新：

也许一些代码可以澄清这个问题。我基本上想做这样的事情：

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(root.get(Entity1_.someProperty), "foo"),   <-- This won't work
        builder.equal(root.get(Entity2_.otherProperty), "bar")   <-- Neither will this
));

Criteria crit = session.createCriteria(YourClassName.class);
Criterion cr1 = Restrictions.lt("propertyname", propertyvalue);
Criterion cr2 = Restrictions.gt("propertyname", propertyvalue);
Criterion cr3 = Restrictions.eq("propertyname", propertyvalue);
Criterion cr4 = Restrictions.eq("propertyname", propertyvalue);
Junction cond = Restrictions.conjunction();
cond.add(cr1).add(cr2).add(cr3).add(cr4)
crit.add(cond);

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;
Root<Entity1> entity1 = builder.treat(root, Entity1.class);
Root<Entity2> entity2 = builder.treat(root, Entity2.class);

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(entity1.get(Entity1_.someProperty), "foo"),
        builder.equal(entity2.get(Entity2_.otherProperty), "bar")
));

CriteriaBuilder=。。。；
CriteriaQuery查询=。。。；
根=。。。；
query.select（root）.where（builder.and(
builder.lessThan（root.get（BaseEntity\ux.someDate），new Date（）），
builder.greaterThan（root.get（BaseEntity_uz.otherDate），new Date（）），
builder.isTrue（root.get（BaseEntity.flag）），
builder.equal（root.get（Entity1.someProperty），“foo”），Try
Junction
。您始终可以执行以下操作：

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(root.get(Entity1_.someProperty), "foo"),   <-- This won't work
        builder.equal(root.get(Entity2_.otherProperty), "bar")   <-- Neither will this
));

Criteria crit = session.createCriteria(YourClassName.class);
Criterion cr1 = Restrictions.lt("propertyname", propertyvalue);
Criterion cr2 = Restrictions.gt("propertyname", propertyvalue);
Criterion cr3 = Restrictions.eq("propertyname", propertyvalue);
Criterion cr4 = Restrictions.eq("propertyname", propertyvalue);
Junction cond = Restrictions.conjunction();
cond.add(cr1).add(cr2).add(cr3).add(cr4)
crit.add(cond);

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;
Root<Entity1> entity1 = builder.treat(root, Entity1.class);
Root<Entity2> entity2 = builder.treat(root, Entity2.class);

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(entity1.get(Entity1_.someProperty), "foo"),
        builder.equal(entity2.get(Entity2_.otherProperty), "bar")
));

我希望我正确地理解了您的问题。
我通过将BaseEntity根向下转换为与子类类型对应的新根来解决这个问题，如下所示：

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(root.get(Entity1_.someProperty), "foo"),   <-- This won't work
        builder.equal(root.get(Entity2_.otherProperty), "bar")   <-- Neither will this
));

Criteria crit = session.createCriteria(YourClassName.class);
Criterion cr1 = Restrictions.lt("propertyname", propertyvalue);
Criterion cr2 = Restrictions.gt("propertyname", propertyvalue);
Criterion cr3 = Restrictions.eq("propertyname", propertyvalue);
Criterion cr4 = Restrictions.eq("propertyname", propertyvalue);
Junction cond = Restrictions.conjunction();
cond.add(cr1).add(cr2).add(cr3).add(cr4)
crit.add(cond);

CriteriaBuilder builder = ...;
CriteriaQuery<BaseEntity> query = ...;
Root<BaseEntity> root = ...;
Root<Entity1> entity1 = builder.treat(root, Entity1.class);
Root<Entity2> entity2 = builder.treat(root, Entity2.class);

query.select(root).where(builder.and(
        builder.lessThan(root.get(BaseEntity_.someDate), new Date()),
        builder.greaterThan(root.get(BaseEntity_.otherDate), new Date()),
        builder.isTrue(root.get(BaseEntity_.flag)),
        builder.equal(entity1.get(Entity1_.someProperty), "foo"),
        builder.equal(entity2.get(Entity2_.otherProperty), "bar")
));

CriteriaBuilder=。。。；
CriteriaQuery查询=。。。；
根=。。。；
Root entity1=builder.treat（Root，entity1.class）；
Root entity2=builder.treat（Root，entity2.class）；
query.select（root）.where（builder.and(
builder.lessThan（root.get（BaseEntity\ux.someDate），new Date（）），
builder.greaterThan（root.get（BaseEntity_uz.otherDate），new Date（）），
builder.isTrue（root.get（BaseEntity.flag）），
builder.equal（entity1.get（entity1\uu1.someProperty），“foo”），
builder.equal（entity2.get（entity2\uu2.otherProperty），“bar”）
));
该解决方案完全基于Hibernate，但我正在寻找一个基于JPA标准API的解决方案。此外，我更愿意使用JPA2元模型来引用属性，而不是字符串标识符。op提出的问题是JPA，而不是Hibernate。不工作它会给出null.someDate和null.otherDate，非法错误它现在工作了，我不知道起初是错误的，但它突然开始工作了