Java 在Hibernate继承中使用联接策略查询超类时如何避免联接子类

共有3类：

 @Entity
    @Inheritance(strategy = InheritanceType.JOINED)
    public class Animal extends BaseEntity {
        String eye;
        public String getEye() {
            return eye;
        }
        public void setEye(String eye) {
            this.eye = eye;
        }
    }

    @Entity
    public class Cat extends Animal {
        String catName;
        public String getCatName() {
            return catName;
        }
        public void setCatName(String catName) {
            this.catName = catName;
        }
    }
    @Entity
    public class Dog extends Animal {
        String dogName;
        public String getDogName() {
            return dogName;
        }
        public void setDogName(String dogName) {
            this.dogName = dogName;
        }
    }

查询动物时，Hibernate生成如下SQL：

SELECT 
  animal0_.id AS id1_7_0_,
  animal0_.eye AS eye2_7_0_,
  animal0_1_.catName AS catName1_8_0_,
  animal0_2_.dogName AS dogName1_9_0_,
  CASE
    WHEN animal0_1_.id IS NOT NULL 
    THEN 1 
    WHEN animal0_2_.id IS NOT NULL 
    THEN 2 
    WHEN animal0_.id IS NOT NULL 
    THEN 0 
  END AS clazz_0_ 
FROM
  Animal animal0_ 
  LEFT OUTER JOIN Cat animal0_1_ 
    ON animal0_.id = animal0_1_.id 
  LEFT OUTER JOIN Dog animal0_2_ 
    ON animal0_.id = animal0_2_.id 
WHERE animal0_.id = ?

就像上面说的

所以，问题是：在查询超类时如何避免加入子类？
通常，只需要超类中的属性——忽略它的子类

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（想象动物有很多子类，如鸟、猪、鱼、昆虫……在这种情况下，动物查询将出现性能问题。）

只选择所需的属性，而不是选择整个实体。