我有两个表。JPA标准加入OneToMany表where子句不起作用
CREATE TABLE public.question
(
id SERIAL PRIMARY KEY
, day VARCHAR(2) NOT NULL
, month VARCHAR(2) NOT NULL
, year VARCHAR(4) NOT NULL
);
CREATE TABLE public.question_translation
(
id SERIAL PRIMARY KEY
, question_id INT REFERENCES public.question(id) NOT NULL
, question_text TEXT NOT NULL
, language VARCHAR(2) NOT NULL
);
现在我想创建条件来检索问题。 SQL是这样的:
SELECT * FROM question q LEFT JOIN question_translation qt ON q.id = qt.question_id WHERE qt.language = 'en'
在Java中使用JPA规定 - 它看起来像这样:
@Override
public Collection<Question> findByMonthYear(String month, String year, String locale) {
EntityManager em = sessionFactory.createEntityManager();
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<Question> criteriaQuery = builder.createQuery(Question.class);
List<Predicate> predicates = new ArrayList<>();
Root<Question> questionRoot = criteriaQuery.from(Question.class);
ListJoin<Question, QuestionTranslation> questionTranslationJoinRoot = questionRoot.join(Question_.questionTranslation, JoinType.LEFT);
predicates.add(builder.equal(questionRoot.get(Question_.month), month));
predicates.add(builder.equal(questionRoot.get(Question_.year), year));
predicates.add(builder.equal(questionTranslationJoinRoot.get(QuestionTranslation_.language), locale));
criteriaQuery.select(questionRoot).where(predicates.toArray(new Predicate[]{}));
TypedQuery<Question> query = em.createQuery(criteriaQuery);
String queryString = query.unwrap(Query.class).getQueryString();
return query.getResultList();
}
我使用ListJoin因为在元模型Question_.class
我得到这一行:
public static volatile ListAttribute<Question, QuestionTranslation> questionTranslation;
但这一个返回我问题类与列表QuestionTranslation两个条目,其中语言字段等于en
和de
值。但是我指定where子句只返回一个条目,其中语言等于en
的值。我的代码有什么问题?
更新#1:
我有第二种情况。
还有一个表:
CREATE TABLE public.user_answer
(
uuid VARCHAR(36) PRIMARY KEY
, user_uuid VARCHAR(36) REFERENCES public.users(uuid) NOT NULL
, question_id INT REFERENCES public.question(id) NOT NULL
, answer TEXT NOT NULL
);
,我想做出这样的SQL:
SELECT * FROM user_answer ua LEFT JOIN question q on ua.question_id = q.id LEFT JOIN question_translation qt ON q.id = qt.question_id WHERE qt.language = 'en' AND ua.user_uuid = '00000000-user-0000-0000-000000000001' AND q.month = '01' AND q.day = '01' AND q.year = '2016';
在Java中使用JPA规定 - 它看起来像这样:
@Override
public UserAnswer findByDayMonthYear(String day, String month, String year, User user, String locale) {
EntityManager em = sessionFactory.createEntityManager();
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<UserAnswer> criteriaQuery = builder.createQuery(UserAnswer.class);
List<Predicate> predicates = new ArrayList<>();
Root<UserAnswer> userAnswerRoot = criteriaQuery.from(UserAnswer.class);
Join<UserAnswer, Question> questionJoin = userAnswerRoot.join(UserAnswer_.question);
ListJoin<Question, QuestionTranslation> questionTranslatJoin = questionJoin.join(Question_.questionTranslation);
predicates.add(builder.equal(builder.treat(questionJoin, Question.class).get(Question_.day), day));
predicates.add(builder.equal(builder.treat(questionJoin, Question.class).get(Question_.month), month));
predicates.add(builder.equal(builder.treat(questionJoin, Question.class).get(Question_.year), year));
predicates.add(builder.equal(builder.treat(questionTranslatJoin, QuestionTranslation.class).get(QuestionTranslation_.language), locale));
predicates.add(builder.equal(userAnswerRoot.get(UserAnswer_.user), user));
criteriaQuery.select(userAnswerRoot).where(predicates.toArray(new Predicate[]{}));
TypedQuery<UserAnswer> query = em.createQuery(criteriaQuery);
String queryString = query.unwrap(Query.class).getQueryString();
return query.getSingleResult();
}
在这种情况下,问题列出了包含两个带有languag的QuestionTranlsation项目es en
和de
,但我只需要一个QuestionTranlsation条目,其语言等于en
。
我在这种情况下要做什么?
看看我的问题更新,请。 – dikkini
为什么我需要它?您可以通过翻译访问问题。 –
我使用JPA 2.1,给我一个建议如何使用'JOIN ON语言'? – dikkini