android - 如何使用 POST 方法在 android 中使用 Retrofit 发送原始 JSON

Question

我想发布以下 JSON，

{
  "name": {
          "firstName": "f_name",
          "lastName": "l_name"
  },
  "password": "mypassword123",
  "email": "test.mail@gmail.com"
}

接口(interface)内部的注册方法是，

@POST("user/createuser")
Call<RegisterResponseModel> register(@Body RegisterModel body);

使用像这样的注册方法，

Name name = new Name("f_name", "l_name");

RegisterModel registerModel = new RegisterModel(name, "mypassword123", "test.mail@gmail.com");

Call<RegisterResponseModel> res = apiService.register(registerModel);

但无法达到我想要的，请帮助我实现我需要的。提前致谢

Answer 1

试试这个

我在尝试仅以 原始 POST 数据 形式时遇到了同样的问题

1. 您的 API 接口(interface)

---

@POST("your_url_here")
Call<Object> getUser(@Body Map<String, String> body);

---

1. 调用您的类

---

Retrofit retrofit = new Retrofit.Builder()
        .baseUrl(Constants.BASE_URL)
        .addConverterFactory(GsonConverterFactory.create())
        .build();
ApiInterface apiInterface = retrofit.create(ApiInterface.class);

try {
    Map<String, String> requestBody = new HashMap<>();
    requestBody.put("email", "abc@gmail.com");
    requestBody.put("password", "123678");
    Call<Object> call=apiInterface.getUser(requestBody);
    call.enqueue(new Callback<Object>() {
        @Override
        public void onResponse(Call<Object> call, Response<Object> response) {
            try {
                JSONObject object=new JSONObject(new Gson().toJson(response.body()));
                Log.e("TAG", "onResponse: "+object );
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
        @Override
        public void onFailure(Call<Object> call, Throwable t) {
        }
    });
} catch (Exception e) {
    e.printStackTrace();
}