java - JPA Criteria 从一些表中选择

Question

表的结构如下:

CREATE TABLE Train
(
    idTrain INT NOT NULL AUTO_INCREMENT,
    nameTrain CHAR(20) NOT NULL,
    PRIMARY KEY (idTrain)
);

CREATE TABLE Station
(
    idStation INT NOT NULL AUTO_INCREMENT,
    nameStation CHAR(50) NOT NULL,
    PRIMARY KEY (idStation)
);

CREATE TABLE Schedule
(
    idStation INT NOT NULL,
    idTrain INT NOT NULL,
    arrivalTime TIME NOT NULL,
    departureTime TIME NOT NULL,
    nextStation INT NOT NULL,
    kmToNextStation INT NOT NULL,
    PRIMARY KEY (idStation, idTrain, nextStation),
    FOREIGN KEY (idStation) REFERENCES Station(idStation),
    FOREIGN KEY (idTrain) REFERENCES Train(idTrain),
    FOREIGN KEY (nextStation) REFERENCES Station(idStation)
);

需要使用 JPA Criteria 实现以下 sql 查询:

SELECT Station.nameStation, Schedule.arrivalTime, Schedule.departureTime, Schedule.kmToNextStation
FROM Schedule
JOIN Station
    ON Station.idStation = Schedule.idStation
JOIN Train
    ON Schedule.idTrain = Train.idTrain
WHERE Train.nameTrain = "268A";

这是我的尝试:

EntityManager em = EntitySupport.getEntityManager();
        CriteriaBuilder builder = em.getCriteriaBuilder();
        CriteriaQuery<ScheduleEntity> cq = builder.createQuery(ScheduleEntity.class);
        Root<ScheduleEntity> root = cq.from(ScheduleEntity.class);
        Join<ScheduleEntity, TrainEntity> idTrain = root.join("idTrain");
        Join<ScheduleEntity, StationEntity> idStation = root.join("idStation");
        cq.multiselect(root.get("arrivalTime"),
                root.get("departureTime"),
                idTrain.get("nameTrain"));
        Query query = em.createQuery(cq);
        List res = query.getResultList();
        System.out.println("Result query: " + res.toString());

显然，我做错了什么，因为我收到以下错误:

Exception in thread "main" java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Unable to locate appropriate constructor on class [srt.entity.ScheduleEntity]. Expected arguments are: java.util.Date, java.util.Date, java.lang.String [select new srt.entity.ScheduleEntity(generatedAlias0.arrivalTime, generatedAlias0.departureTime, generatedAlias1.nameTrain) from srt.entity.ScheduleEntity as generatedAlias0 inner join generatedAlias0.idTrain as generatedAlias1 inner join generatedAlias0.idStation as generatedAlias2]

帮助我使用 JPA Criteria 为上述 sql 查询编写正确的代码。

Answer 1

我对 JPA Criteria 的概念是错误的。现在我完全明白了，答案如下:

EntityManager em = EntitySupport.getEntityManager();
        CriteriaBuilder builder = em.getCriteriaBuilder();
        CriteriaQuery<ScheduleEntity> cq = builder.createQuery(ScheduleEntity.class);
        Root<ScheduleEntity> from = cq.from(ScheduleEntity.class);
        Join<ScheduleEntity, StationEntity> idStation = from.join("idStation");
        Join<ScheduleEntity, TrainEntity> idTrain = from.join("idTrain");
        Predicate where = builder.equal(idTrain.get("nameTrain"), "268A");
        cq.where(where);
        List<ScheduleEntity> schedule = em.createQuery(cq).getResultList();
        for(ScheduleEntity s : schedule) {
            System.out.println(s.toString());
        }