如何反序列化可以是字符串，对象或列表的json / gson

一尘不染

如何反序列化可以是字符串，对象或列表的json / gson

json

我有以下json

"notes": {"note": [
         {
             "content": "Having wisdom teeth removed.",
             "from": "employee"
         },
         {
             "content": "Get well soon",
             "from": "manager"
         }
     ]},

问题在于，价值价值也是

 "notes": "",

要么

"notes": {"note": {
            "content": "This is a test note.",
            "from": "employee"
        }},

并将其存储在这些

public  class Notes
{
    @SerializedName ("note")
    public List<Note> note;
}
public  class Note
{
    @SerializedName ("content")
    public String content;
    @SerializedName ("from")
    public String from;
}

我相信我通过这样做解决了不是数组而是单个对象的问题

public class Json {
    private static Gson gson;

    private static class MyNoteClassTypeAdapter implements JsonDeserializer<List<RequestsDTO.Note>> {
        public List<RequestsDTO.Note> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext ctx) {
            List<RequestsDTO.Note> vals = new ArrayList<RequestsDTO.Note>();
            if (json.isJsonArray()) {
                for (JsonElement e : json.getAsJsonArray()) {
                    vals.add((RequestsDTO.Note) ctx.deserialize(e, RequestsDTO.Note.class));
                }
            } else if (json.isJsonObject()) {
                vals.add((RequestsDTO.Note) ctx.deserialize(json,RequestsDTO.Note.class));
            } else {
                throw new RuntimeException("Unexpected JSON type: " + json.getClass());
            }
            return vals;
        }
    }

    public static Gson getGson()
    {
        if (gson == null)
        {
            Type ListType = new TypeToken<List<RequestsDTO.Note>>() {}.getType();
            GsonBuilder builder = new GsonBuilder();
            builder.registerTypeAdapter(DateTime.class, new DateTimeSerializer());
            builder.registerTypeAdapter(ListType, new MyNoteClassTypeAdapter());
            gson = builder.create();
        }
        return gson;
    }
}

现在，当整个事情以字符串形式返回时，我仍然处于困境....

阅读 230

2020-07-27

共1个答案

一尘不染

我们的想法是试图让"note"（从外地"notes"
JSONObject）为JSONArray第一，如果抛出异常，这将意味着，没有"note" JSONArray进入"notes"
JSONObject ，这将意味着"note"是JSONObject。用同样的方法，我们可以计算出状况时note场String。

try {
        //String jsonString="{\"notes\": {\"note\": [{\"content\": \"Having wisdom teeth removed.\",\"from\": \"employee\" }, {\"content\": \"Get well soon\", \"from\": \"manager\"} ] }}";
        //String jsonString="{\"notes\": { \"note\": {\"content\": \"This is a test note.\",\"from\": \"employee\"}}}";
        String jsonString="{\"notes\": { \"note\": \"\"}}";

        JSONObject jsonObject=new JSONObject(jsonString);
        JSONObject jsonObjectNotes=jsonObject.getJSONObject("notes");

        try{
            JSONArray jsonArrayNote=jsonObjectNotes.getJSONArray("note");
            for (int i = 0; i < jsonArrayNote.length(); i++) {

                JSONObject jsonObject2= jsonArrayNote.getJSONObject(i);
                String stringContent=jsonObject2.getString( "content");
                String stringFrom= jsonObject2.getString( "from");

                Log.e(getClass().getName(), "content="+stringContent +"; from="+stringFrom);
            }
        }
        catch(JSONException e){
            //that means that jsonObjectNotes has no jsonArray with name "notes" and "notes" is jsonObject
            try{
                JSONObject jsonObject3=jsonObjectNotes.getJSONObject("note");

                String stringContent=(String) jsonObject3.get( "content");
                String stringFrom=(String) jsonObject3.get( "from");

                Log.e(getClass().getName(), "content="+stringContent +"; from="+stringFrom);
            }
            catch(JSONException ex){
                //that means that jsonObjectNotes has no jsonObject with name "notes" and "notes" is empty String
                String stringNote=jsonObjectNotes.getString("note") ;       
                Log.e(getClass().getName(), "note is string ="+ stringNote);
            }
        }

    } catch (JSONException e) {
        e.printStackTrace();
    }

在我的示例代码中，其他get操作也可能引发jsonExceptions，但是我想您已经明白了。

2020-07-27