7

我的实体类如下

@Entity
@table
public class User {

    @OneToOne
    private UserProfile userProfile;

    // others
}


@Entity
@Table
public class UserProfile {

    @OneToOne
    private Country country;
}

@Entity
@Table
public class Country {
    @OneToMany
    private List<Region> regions;
}

现在我想获取特定区域的所有用户。我知道 sql 但我想通过 spring data jpa Specification 来做到这一点。以下代码不应该工作,因为区域是一个列表,我正在尝试与单个值匹配。如何获取区域列表并与单个对象进行比较?

 public static Specification<User> userFilterByRegion(String region){


        return new Specification<User>() {
            @Override
            public Predicate toPredicate(Root<User> root, CriteriaQuery<?> criteriaQuery, CriteriaBuilder criteriaBuilder) {

                return criteriaBuilder.equal(root.get("userProfile").get("country").get("regions").get("name"), regionalEntity);
            }
        };
    }

编辑:感谢您的帮助。实际上我正在寻找以下 JPQL 的等效条件查询

SELECT u FROM User u JOIN FETCH u.userProfile.country.regions ur WHERE ur.name=:<region_name>
4

2 回答 2

7

尝试这个。这应该工作

criteriaBuilder.isMember(regionalEntity, root.get("userProfile").get("country").get("regions"))

Equals您可以通过覆盖Region 类中的方法(也是 Hashcode)来定义相等的条件

于 2017-12-18T11:14:27.600 回答
1

我的代码片段

// string constants make maintenance easier if they are mentioned in several lines
private static final String CONST_CLIENT = "client";
private static final String CONST_CLIENT_TYPE = "clientType";
private static final String CONST_ID = "id";
private static final String CONST_POST_OFFICE = "postOffice";
private static final String CONST_INDEX = "index";
...

@Override
public Predicate toPredicate(Root<Claim> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
    List<Predicate> predicates = new ArrayList<Predicate>();
    // we get list of clients and compare client's type
    predicates.add(cb.equal(root
                .<Client>get(CONST_CLIENT)
                .<ClientType>get(CONST_CLIENT_TYPE)
                .<Long>get(CONST_ID), clientTypeId));
    // Set<String> indexes = new HashSet<>();
    predicates.add(root
                .<PostOffice>get(CONST_POST_OFFICE)
                .<String>get(CONST_INDEX).in(indexes));
    // more predicates added
    return return andTogether(predicates, cb);
}

private Predicate andTogether(List<Predicate> predicates, CriteriaBuilder cb) {
    return cb.and(predicates.toArray(new Predicate[0]));
}

如果您确定只需要一个谓词,则使用 List 可能是一种矫枉过正。

于 2017-12-18T11:25:11.097 回答