Question

我正在尝试创建Photo对象的Photos ArrayList。我想将Photo类与Photos类分开，因为它变得不可读。问题是我不再收到任何数据

值得注意的是，当我将Photo类嵌套在Photos类中时，这个WAS工作如下：

public class Photos { 
    @Expose private int page;
    @Expose private int pages;
    @Expose private int perpage;
    @Expose private String total;
    @Expose private ArrayList<Photo> photo = new ArrayList<Photo>();

    //helpers 
    public ArrayList<Photo> getPhotos(){return photo;}
    public void setPhotos(ArrayList<Photo> photos){this.photo = photos;}

//want to put this in it's own class
public class Photo { 
    //helpers 
    @Override 
    public String toString(){
        return title;
    } 
    public String getUrl(){return url_s;}

    //GSON fields 
    @Expose private String id;
    @Expose private String owner;
    @Expose private String secret;
    @Expose private String server;
    @Expose private int farm;
    @Expose private String title;
    @Expose private int ispublic;
    @Expose private int isfriend;
    @Expose private int isfamily;
    @Expose private String url_s;
}

}

改造：

 new Callback<PhotosResponse>() {
                    @Override
                    public void success(PhotosResponse photosResponse, Response response) {
                    bus.post(new ImagesReceivedEvent(photosResponse.getPhotosObject().getPhotos()));
                }

参数与JSON完全匹配，我通过以下响应收到了一个arraylist：

public class ImagesReceivedEvent {

    private ArrayList<Photo> result;

    public ImagesReceivedEvent(ArrayList<Photo> result){
        Log.i(TAG, "arraylist for presenter");
        this.result = result;
    }

    public ArrayList<Photo> getResult(){return result;}
}

这是收到回复的地方，但不再是：

public class PhotosResponse {

    @Expose private Photos photos;
    @Expose private String stat;

    public Photos getPhotosObject(){return photos;}

    public void setPhotosObject(Photos photos) {
        this.photos = photos;
    }

    public String getStat() {
        return stat;
    }

    public void setStat(String stat) {
        this.stat = stat;
    }
}

如何将这些类分开并仍将arrayList作为响应？我想将它们分开，因为我计划在每个类上使用Parcelable，在使用内部类时我需要跳过一些障碍。任何指导将不胜感激！感谢。

编辑：这是JSON有效负载：

{ "photos": { "page": 1, "pages": 10, "perpage": 100, "total": "1000", 
"photo": [
  { "id": "18473086614", "owner": "130897025@N07", "secret": "b3c684c356", "server": "259", "farm": 1, "title": "My travels circa 2013. Summer days on #charlesbridge #charlesbridgeprague #prauge. upbound@upbound.net || 718-754-5850 #photographer #photography #canonphotography #canon #canondslr #canon600d #dslr #600d #travelphotography #europe #upboundonline #canonre", "ispublic": 1, "isfriend": 0, "isfamily": 0 },
  { "id": "18473090744", "owner": "131790787@N07", "secret": "2734055852", "server": "3705", "farm": 4, "title": "Untitled", "ispublic": 1, "isfriend": 0, "isfamily": 0 },
  { "id": "18473091934", "owner": "61308696@N00", "secret": "b40dbfcf15", "server": "401", "farm": 1, "title": "Climbing Acatenango", "ispublic": 1, "isfriend": 0, "isfamily": 0 },
  { "id": "18473092714", "owner": "39055811@N08", "secret": "e51f5a183b", "server": "3936", "farm": 4, "title": "DSCF1735.jpg", "ispublic": 1, "isfriend": 0, "isfamily": 0 }
] }, "stat": "ok" }

Answer 1

事实证明我在命名约定上犯了一个错误：

@Expose private ArrayList<Photo> mPhotos = new ArrayList<Photo>();

我已将照片重命名为arrayList上的mPhotos，但忘记了GSON需要JSON字段的精确名称（原来是照片）。我把它改回了＆＃34;照片＆＃34;现在，arrayList填充在我的Photos对象上。

我需要向自己强调所有字段都必须符合JSON命名约定。

感谢。

Answer 2

当使用Retrofit和GSON时，无法将您的响应“分离”到不同的类中，您必须自己以后再进行。使用所有信息接收您的大对象，然后将其打开到所需的类。

在我做过的项目中，我有一个包含Wrapper类的特定包来获取Retrofit的响应，然后我会手动将它映射到我实际需要的对象。希望它有所帮助。

改造：使用GSON将JSON解析为多个Java对象

2 个答案: