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问题描述

我正在创建一个具有2个一对多关系的实体.事件具有用户字段和位置字段.我正在尝试使用自动查询,但是此代码始终返回一个空列表.

I'm creating a an entity with 2 one-to-many relationships. An Event has a User field and a Place field. I'm trying to use the automatic-query, but this code always returns an empty list.

    User user = new User("[email protected]","Maurizio Pozzobon","01","hash","facebook");
    user.insert();
    Place place = new Place("posto","bel posto",null,null);
    place.insert();
    Event e =new Event(user,place, "Festa","Questa è una gran bella festa",null,new Date(),(long) 10,false,null);
    e.insert();
    List<Event> l =user.events.fetch();

事件类是

public class Event extends Model{
    @Id
    public Long id;

    ...

    //Relazioni
    public User user;
    public Place place;

        ...

    public Event(User user, Place place,String nome, String descrizione, String uRLImmagine, Date dataInizio, Long durata, Boolean isRicorrente, Long ricorrenza) {
        this.user=user;
        this.place=place;
        ...
    }
        ...
}

如果我这样更改Event类

If I change the Event class like this

public class Event extends Model{
    @Id
    public Long id;

    ...

    //Relazioni
    public User user;
    //public Place place;

        ...

    public Event(User user, Place place,String nome, String descrizione, String uRLImmagine, Date dataInizio, Long durata, Boolean isRicorrente, Long ricorrenza) {
        this.user=user;
        //this.place=place;
        ...
    }
        ...
}

上面的相同代码返回一个包含一个事件的列表(我所期望的)

The same code above returns a list with one Event in it (what I expected)

这是Place类

public class Place extends Model {



@Id
public Long id;

public String nome;
public String descrizione;
public String uRLImmagine;
public String indirizzo;


//Relazioni
// public User user;
//@Filter("place")
//public Query<Event> events;
private Set<Long> idEvents = new HashSet<Long>();
private Set<Long> idPlaceVotes = new HashSet<Long>();
private Set<Long> idPlaceComments = new HashSet<Long>();


public Place(/*User user,*/ String nome, String descrizione, String uRLImmagine,String indirizzo) {
//  this.user=user;
    this.nome = nome;
    this.descrizione = descrizione;
    this.uRLImmagine = uRLImmagine;
    this.indirizzo = indirizzo;
}


static Query<Place> all() {
    return Model.all(Place.class);
}

public static Place findById(Long id) {
    return all().filter("id", id).get();
}

public String toString() {
    return nome;
}  

public static void delete(Long id) {
    findById(id).delete();

}

}

这是User类

public class User extends Model {

@Id
public Long id;

public String nome;
public String email;
public String webId;    //ID of the user in the provider website
public String passwordHash;
public String service;

//Relazioni
@Filter("user")
public Query<Event> events;

public User(String email, String name,String webId, String passwordHash, String service) throws Exception {
    if (email!=null)
        this.email=email;
    else
        throw new Exception("An email is required");
    if (name!=null)
        this.nome=name;
    else
        throw new Exception("A name is required");

    if (webId!=null)
        this.webId=webId;
    else
        throw new Exception("A webId is required");

    this.passwordHash=passwordHash;
    this.service = service;
}

public void setEmail(String email) throws Exception{
    if (email!=null)
        this.email=email;
    else
        throw new Exception("An email is needed");
}

static Query<User> all() {
    return Model.all(User.class);
}

public static User findById(Long id) {
    return all().filter("id", id).get();
}

public static User findByEmail(String email){
    return all().filter("email", email).get();
}

public String toString() {
    return nome;
}

}

MyModel是siena.Model的超类,但是正确地知道它没有任何用处,因此我将其改回了Model.我在游戏1.1上使用play-siena 1.5

MyModel was a super class of siena.Model, but right know it doesn't do anything useful so I changed it back to Model.I'm using play-siena 1.5 on play 1.1

推荐答案

我发现了您的问题:)

这是一个已知问题,但我每次都忘记它.
在您的事件中,只需声明@Column:

This is a known issue but I forget it each time.
In your event, simply declare the @Column:

public class Event extends Model{
    @Id
    public Long id;

    @Column("user")
    public User user;

    @Column("place")
    public Place place;
}

由于User和Place分别具有一个名为"id"的键字段,并且在没有@Column的情况下,Siena默认使用键字段名,因此当GAE尝试通过字段"id"查找对象时会发生冲突.
我将尝试在Siena v1.0.0中更正此问题(并且您已经可以在Play中透明地使用从v1.0.0中继生成的siena.jar)

As User and Place have each a key field named "id" and the key field name is used by default by Siena when there is no @Column, there is a collision when GAE tries to find the object by field "id".
I will try to correct this in Siena v1.0.0 (and you can already use siena.jar generated from v1.0.0 trunk transparently in Play)

致谢帕斯卡

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09-26 16:47