Question

我的JSON响应具有以下结构。我应该如何解析它以保存结构存储在java中的列表中。 GSON是否提供了解决方案，还是应该使用多维数组列表？

我需要将每个标记数组传递给不同的视图分页器，我应该遵循什么方法。

{
  "tag": [
    {
      "listing_count": 5,
      "listings": [
        {
          "source": "source1",
          "data": {
            "image": "image1",
            "name": "name1"
          },
          "name": "name1"
        }
      ]
    },
    {
      "listing_count": 5,
      "listings": [
        {
          "source": "source2",
          "data": {
            "image": "imag2",
            "name": "name2"
          },
          "name": "name2"
        }
      ]
    }
  ]
}

修改

我已按照答案中的建议创建了类。我在创建GSON响应类时遇到问题。这就是我创造的：

public GsonRequest(int method, String url, Class<T> clazz,
            Listener<T> listener, ErrorListener errorListener, Gson gson) {
        super(Method.GET, url, errorListener);
        this.mClazz = clazz;
        this.mListener = listener;
        mGson = gson;
    }

Answer 1

gson类应该是这样的：

public class TagList {

ArrayList<Tag> tags;

public static class Tag {
    int listing_count;
    ArrayList<Listings> listings;

    public int getListing_count() {
        return listing_count;
    }

    public void setListing_count(int listing_count) {
        this.listing_count = listing_count;
    }

    public ArrayList<Listings> getListings() {
        return listings;
    }

    public void setListings(ArrayList<Listings> listings) {
        this.listings = listings;
    }

}

public static class Listings {
    String source;
    Data data;
    String name;

    public String getSource() {
        return source;
    }

    public void setSource(String source) {
        this.source = source;
    }

    public Data getData() {
        return data;
    }

    public void setData(Data data) {
        this.data = data;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

public static class Data {
    String image;
    String name;

    public String getImage() {
        return image;
    }

    public void setImage(String image) {
        this.image = image;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

在android中解析并存储json响应

1 个答案: