java - SPRING DATA JPA保存@OneToMany关系

我尝试以一对多关系保留父实体:

@Entity
public class TrainEx  {
    private Set<TrainCompositionEx> trainCompositionsByTrainId;
    @OneToMany(cascade = {CascadeType.ALL}, fetch = FetchType.LAZY,  mappedBy = "trainByTrainId")
    public Set<TrainCompositionEx> getTrainCompositionsByTrainId() {
        return trainCompositionsByTrainId;
    }

    public void setTrainCompositionsByTrainId(Set<TrainCompositionEx> trainCompositionsByTrainId) {
        this.trainCompositionsByTrainId = trainCompositionsByTrainId;
    }
...
}

和子实体:

@Entity
public class TrainCompositionEx{
    @Id
    @ManyToOne(cascade = {CascadeType.REFRESH}, fetch = FetchType.LAZY)
    @JoinColumn(name = "trainId", referencedColumnName = "trainId", nullable = false, insertable = true, updatable = true)
    private TrainEx trainByTrainId;
....

}

所以我从json POST端点收到了TrainEx trainEx:

 @RequestMapping(method= RequestMethod.POST, consumes = "application/json", produces = "application/json")
    public @ResponseBody
    ResponseEntity<Void> addTrain(@RequestBody TrainEx trainEx) throws Exception {
        trainService.add(trainEx);
        return new ResponseEntity<Void>(HttpStatus.CREATED);
    }

json:

 {
    "trainId" : 5,
    "status" :  1,
    "maxWeight" :  200,
    "maxLength" :  35,
    "speed" :  60,
    "totalWeight" : 100,
    "totalLength" : 20,
    "trainCompositionsByTrainId": [{
        "wagonByWagonId": {"wagonId" : 2}
    }]
 }

在我这样保存之后:

...
@Transactional
    public TrainEx add(TrainEx trainEx) {

    for(TrainCompositionEx trainCompositionEx : trainEx.getTrainCompositionsByTrainId()){
        trainCompositionEx.setTrainByTrainId(trainEx);
        trainCompositionEx.setWagonByWagonId(
                em.getReference(WagonEx.class, trainCompositionEx.getWagonByWagonId().getWagonId()));
    }
    return trainExRepository.save(trainEx);
    }
    ...

但是我收到了SQL错误:“trainid”列中的null值违反了非null约束，但是如您所见，我将trainEx实体设置为TrainCompositionEx，并且在调试模式下停止运行，并且trainId在那里存在:

所以我该怎么做？

UPDATE1:
我调查了日志，认为那个孩子的问题在父实体之前仍然存在，因为我插入了train_composition表，但没有插入train表，请参见:

Hibernate: insert into tms.public.train_composition (version, transportOrderId, wagonId, trainId) values (?, ?, ?, ?)

最佳答案

我使用em.persist持久化类。我的上级实体有

@JsonProperty(TIMES)
@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER, orphanRemoval = true)
private Set<Times> timesList = new HashSet<Times>();

当孩子有

@JsonIgnore
@ManyToOne(cascade = CascadeType.ALL)
private Rider rider;

该代码是

em.persist(rider);
for (Times t : rider.getTimes()) {
    t.setRider(rider);
}

对于新手，对于现有手，我将新数据合并到旧对象中。
该项目在这里:https://github.com/xtien/motogymkhana-server
该代码来自RiderDaoImpl，Rider和Times。
这是项目的目的:http://www.gymcomp.com/eu

关于java - SPRING DATA JPA保存@OneToMany关系，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/36163310/