java - JSON获取没有对象的数组

Question

我正在尝试解析 this JSON ,

{
  "listname": "red",
  "lists": [
    {
      "id": "01",
      "name": "paw",
      "list": [
        {
          "id": "A",
          "name": "pawa",
          "bar": "foo"
        },
        {
          "id": "B",
          "name": "pawb",
          "bar": "foo"
        }
      ]
    },
    {
      "id": "02",
      "name": "pew",
      "list": [
        {
          "id": "A",
          "name": "pewa",
          "bar": "foo"
        },
        {
          "id": "B",
          "name": "pewb",
          "bar": "foo"
        }
      ]
    },
    {
      "id": "03",
      "name": "piw",
      "list": [
        {
          "id": "A",
          "name": "piwa",
          "bar": "foo"
        },
        {
          "id": "B",
          "name": "piwb",
          "bar": "foo"
        }
      ]
    }
  ]
}

我把它放在 Asset Folder 上，我读了它，它把它转换成 String 因为这里一切都很好，问题是当我试图得到lists 中的每个项目的 name 并尝试从 list 中获取所有名称 我试过这样做:

JSONObject obj = new JSONObject(str);
JSONArray jsonMainArr = obj.getJSONArray("lists"); //first get the lists
for (int i = 0; i < jsonMainArr.length(); i++) { 
   JSONObject childJSONObject = jsonMainArr.getJSONObject(i);
   String name = childJSONObject.getString("name");
   Toast.makeText(context, name, Toast.LENGTH_SHORT).show();
}

JSONObject obj = new JSONObject(str);
JSONArray jsonMainArr = obj.getJSONArray("list"); //get the list
for (int i = 0; i < jsonMainArr.length(); i++) { 
   JSONObject childJSONObject = jsonMainArr.getJSONObject(i);
   String name = childJSONObject.getString("name");
   Toast.makeText(context, name, Toast.LENGTH_SHORT).show();
}

但它没有显示任何内容……我错过了什么？

编辑

这就是我读取 JSON

的方式

 public static String loadJSONFromAsset(Context ctx, String str) {
    String json = null;
    try {

        InputStream is = ctx.getAssets().open(str);

        int size = is.available();
        byte[] buffer = new byte[size];
        is.read(buffer);
        is.close();
        json = new String(buffer, "UTF-8");

    } catch (IOException ex) {
        ex.printStackTrace();
        return null;
    }
    return json;

}

Answer 1

由于您的“列表”嵌套在 childJSONObject(来自“列表”)中，嵌套您的 for 循环以检索这组值(JSONobject)

JSONObject obj = new JSONObject(str);
//first get the top-level "lists"
JSONArray jsonMainArr = obj.getJSONArray("lists");
for (int i = 0; i < jsonMainArr.length(); i++) { 
   JSONObject childJSONObject = jsonMainArr.getJSONObject(i);
   String name = childJSONObject.getString("name");
   Toast.makeText(context, name, Toast.LENGTH_SHORT).show();
   //get the inner "list" from childJSONObject
   JSONArray childJSONArray = childJSONObject.getJSONArray("list"); 
   for (int j = 0; j < childJSONArray.length(); j++) {
      JSONObject childJSONObjectB = childJSONArray.getJSONObject(j);
      String name = childJSONObjectB.getString("name");
      Toast.makeText(context, name, Toast.LENGTH_SHORT).show();
   }
}