如何在Spring Data JPA中对and和OR进行分组？

import org.springframework.data.jpa.domain.Specification;
import javax.persistence.criteria.CriteriaBuilder;
import javax.persistence.criteria.CriteriaQuery;
import javax.persistence.criteria.Predicate;
import javax.persistence.criteria.Root;

// 定义一个Specification
public class AndOrSpecification<T> implements Specification<T> {

    private Predicate andPredicate;
    private Predicate orPredicate;

    public AndOrSpecification(Predicate andPredicate, Predicate orPredicate) {
        this.andPredicate = andPredicate;
        this.orPredicate = orPredicate;
    }

    @Override
    public Predicate toPredicate(Root<T> root, CriteriaQuery<?> query, CriteriaBuilder criteriaBuilder) {
        return criteriaBuilder.and(andPredicate, criteriaBuilder.or(orPredicate));
    }
}

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.JpaSpecificationExecutor;

public interface UserRepository extends JpaRepository<User, Long>, JpaSpecificationExecutor<User> {
}

import org.springframework.data.jpa.domain.Specification;

public class UserSpecifications {

    public static Specification<User> withAgeAndName(String name, int age) {
        return (root, query, criteriaBuilder) ->
                criteriaBuilder.and(
                        criteriaBuilder.equal(root.get("name"), name),
                        criteriaBuilder.or(
                                criteriaBuilder.greaterThan(root.get("age"), age),
                                criteriaBuilder.isNull(root.get("age"))
                        )
                );
    }
}

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

import java.util.List;

@Service
public class UserService {

    private UserRepository userRepository;

    @Autowired
    public UserService(UserRepository userRepository) {
        this.userRepository = userRepository;
    }

    public List<User> getUsersWithAgeAndName(String name, int age) {
        Specification<User> spec = UserSpecifications.withAgeAndName(name, age);
        return userRepository.findAll(spec);
    }
}