Question

我正在尝试通过retrofit使用登录api。我只需要发送手机号码。当我使用postman主体时，它正在输出。但是当iam用android呼叫时，会收到如下的错误json

{
"error": "Validation error",
"error_code": "001",
"Validation_errors": {
    "mobile": "<p>The Mobile field is required.</p>"
}
}

HomeActivity.class

 ApiInterface apiService =
                                ApiClient.getClient().create(ApiInterface.class);
                        Map<String,String> user = new HashMap<>();
                        user.put("mobile",username.getText().toString().trim());

                        Call<ResponseBody> mService = apiService.loginwithno(user);
                        Log.d("TAG", "response: " + mService.toString());
                        mService.enqueue(new Callback<ResponseBody>() {
                            @Override
                            public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                                if (response.isSuccessful()) {
                                    try {
                                        String result = response.body().string();
                                        JSONObject mJsonObject = new JSONObject(result);
                                        Log.d("TAG", "response: " + mJsonObject.toString());

                                    } catch (JSONException e) {
                                        e.printStackTrace();
                                    } catch (IOException e) {
                                        e.printStackTrace();
                                    }
                                } else {
                                    buttonVisible();
                                    username.setError("Please try again");
                                }
                            }

                            @Override
                            public void onFailure(Call<ResponseBody> call, Throwable t) {
                                call.cancel();
                                buttonVisible();
                                Snackbar snackbar = Snackbar.make(buttonLogin,
                                        "Please check your internet connection", Snackbar.LENGTH_LONG);
                                snackbar.show();

                            }

ApiClient

public class ApiClient {
public static final String BASE_URL = "http://nast.in/driverpool/api/index.php/";
private static Retrofit retrofit = null;

public static Retrofit getClient() {
        retrofit = new Retrofit.Builder()
                .baseUrl(BASE_URL)
                .addConverterFactory(GsonConverterFactory.create())
                .build();
    return retrofit;
}
}

ApiInterface

public interface ApiInterface {


@POST("account/login?")
Call<ResponseBody> loginwithno(@Body Map<String, String> mobile);

@POST("account/verifyotp")
Call<ResponseBody> verifyotp(@Body HashMap<String, String> mobile);//Param name: mobile, otp

@POST("account/resendotp")
Call<ResponseBody> resentotp(@Body HashMap<String, String> mobile);
}

邮递员屏幕截图

Answer 1

首先，您不需要'？'在您的api中，我认为您必须在@body中发送json，因此创建此类

npm WARN tar invalid entry

并在ApiInterface中使用它

public class SendLoginData{
    @SerializedName("mobile")
    public String mobile;

    public SendLoginData(String mobile) {
        this.mobile = mobile;
    }
}

Answer 2

您需要对代码进行少量更改。

更改登录api以接收此类json，如果尚未在项目中添加，请包含gson库。
```
@POST("account/login?")
Call loginwithno(@Body Map mobile);
```

创建一个ApiErrorResponse对象来处理您的api错误。根据需要添加getter，setter和@SerializedName。


 class ApiErrorResponse{
    String error;
    String error_code;
    ValidationErrors Validation_errors; 
 } 
 class ValidationErrors{
    String mobile;
 }

关于API错误句柄


   if(!response.isSuccessful()){
        Converter converter =
                ApiClient.getClient().responseBodyConverter(ApiErrorResponse.class, new Annotation[0]);
        ApiErrorResponse errors = null;
        try {
            errors = converter.convert(response.errorBody());
        } catch (Exception e) {

    }
    if(errors!=null){
     //Handle your API Error logic here
    }


    }
    if(errors!=null){
     //Handle your API Error logic here
    }

在Android中进行改造登录

2 个答案: