java - Gson 解析 JSON，如何要求所有 field 存在于 JSON 中？

Answer 1

Just use reflection to detect whether json has this field.

class A {
    String a;
    String b;
    List<B> list;
}

class B{...}

private static void checkJson(JSONObject json, Class<?> clazz) {
    Field[] fields = clazz.getDeclaredFields();

    for (Field field : fields) {
        if (!json.has(field.getName())) {
            throw new JsonParseException();
        }

        // 遍历数组
        Type type = field.getGenericType();

        if (type instanceof ParameterizedType) {
            Type[] types = ((ParameterizedType) type).getActualTypeArguments();
            Type t = types[0];

            JSONArray array = json.getJSONArray(field.getName());

            for (int i = 0; i < array.length(); i++) {
                Object childJSON = array.get(i);

                if (childJSON instanceof JSONObject) {
                    checkJson((JSONObject) childJSON, (Class) t);
                }
            }
        }
    }
}

public static void main(String[] args){
    A a = new A();
    ...
    String json = new Gson().toJson(a);
    checkJson(new JSONObject(json), A.class);
}

Answer 2

It seems that it is not found. You can control it by yourself through business logic. Just make a null judgment on the Object obtained after serialization

Answer 3

You can only write a custom Deserializer. You can refer to stackoverflow. Someone has asked the same question.

http://stackoverflow.com/questions/3163193/strict-json-parsing-with-googles-gson