在 Android Java 中使用用户名和密码传递时，如何使用不记名令牌对 api 进行身份验证？

Question

我正在使用 Retrofit 库通过 API 向服务器发出 post 请求，发送用户名和密码并获得 session 令牌作为响应。我正在使用此令牌（在 header 中作为承载令牌）向具有纬度和经度的服务器创建 POST 以获取数据，但该方法在 enqueue() 的 onFailure 方法中重复结束.

@POST("api/auth/login/username/{username}/password/{password}")
Call<ResponseBody> login(@Path("username")String username, @Path("password")String password);

@POST("api/products/get")
Call<ResponseBody> getData(@Header("Authorization") String authorization);Call<ResponseBody>

在 MainActivity 中：

Retrofit.Builder builder=new Retrofit.Builder()
                           .baseUrl("fake url")
                           .addConverterFactory(GsonConverterFactory.create());
Retrofit retrofit=builder.build();
DataApi companyApi=retrofit.create(DataApi.class);
private String Token="";tokenCall=comapnyApi.login("data","!fakedata");
        tokenCall.enqueue(new Callback<ResponseBody>() {
            @Override
            public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                if (response.isSuccessful()){
                    try {
                        Toast.makeText(MainActivity.this, response.body().string(), Toast.LENGTH_SHORT).show();
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }
                else {
                    Toast.makeText(MainActivity.this, "Wrong Username and password", Toast.LENGTH_SHORT).show();
                }
            }
            @Override
            public void onFailure(Call<ResponseBody> call, Throwable t) {
                Toast.makeText(MainActivity.this, "Couldn't connect to Fake company Server Api", Toast.LENGTH_SHORT).show();
            }
        });
    Call<ResponseBody> Data=companyApi.getData(Token);
    Data.enqueue(new Callback<ResponseBody>() {
        @Override
        public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
            if (response.isSuccessful()){
                try {
                    Toast.makeText(MainActivity.this, response.body().string(), Toast.LENGTH_SHORT).show();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            else {
                Toast.makeText(MainActivity.this, "Api Token Expired", Toast.LENGTH_SHORT).show();
            }
        }

        @Override
        public void onFailure(Call<ResponseBody> call, Throwable t) {
            Toast.makeText(MainActivity.this, "Couldn't connect to Fake Company Server", Toast.LENGTH_SHORT).show();
        }
    });

Answer 1

我认为这是因为您在第一个请求之后立即提出了第二个请求。 enqueue 是异步的，所以当你第一次请求时，我的意思是 tokenCall.enqueue()，它的响应稍后出现，所以第二个 api 调用或 Data.enqueue() 是用 token == "" 调用的.因此，在第一个 api 调用的 onResponse 中进行第二个 api 调用。

private String Token="";
tokenCall=comapnyApi.login("data","!fakedata");
    tokenCall.enqueue(new Callback<ResponseBody>() {
        @Override
        public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
            if (response.isSuccessful()){
                try {
                    Toast.makeText(MainActivity.this, response.body().string(), Toast.LENGTH_SHORT).show();
                    Token = response.body().string();
                    Call<ResponseBody> Data=companyApi.getData(Token);
                    Data.enqueue(new Callback<ResponseBody>() {
                        @Override
                        public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                            if (response.isSuccessful()){
                                try {
                                    Toast.makeText(MainActivity.this, response.body().string(), Toast.LENGTH_SHORT).show();
                                } catch (IOException e) {
                                    e.printStackTrace();
                                }
                            }
                            else {
                                Toast.makeText(MainActivity.this, "Api Token Expired", Toast.LENGTH_SHORT).show();
                            }
        }

        @Override
        public void onFailure(Call<ResponseBody> call, Throwable t) {
            Toast.makeText(MainActivity.this, "Couldn't connect to Fake Company Server", Toast.LENGTH_SHORT).show();
        }
    });
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            else {
                Toast.makeText(MainActivity.this, "Wrong Username and password", Toast.LENGTH_SHORT).show();
            }
        }
            @Override
            public void onFailure(Call<ResponseBody> call, Throwable t) {
                Toast.makeText(MainActivity.this, "Couldn't connect to Fake company Server Api", Toast.LENGTH_SHORT).show();
            }
        });

在 Android Java 中使用用户名和密码传递时，如何使用不记名令牌对 api 进行身份验证？

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json

retrofit