Question

我正在尝试通过改造来检查服务器中是否存在某些数据。我的问题是，无论是onResponse还是OnFailure。代码永远不会传递这两种方法。我在日志中没有收到任何错误。如果我调试，代码停在这里：call.enqueue（new Callback（），并在下一步中走出方法。

一些帮助将不胜感激

代码：

 ApiInterface apiService = ApiClient.getClient().create(ApiInterface.class);

        Call<User> call = apiService.smsPincodeCheck(getPhoneNuber, email);
        call.enqueue(new Callback<User>() {
            @Override
            public void onResponse(Call<User> call, Response<User> response) {
                if(response.isSuccess()){
                    Toast.makeText(RegisterThreeActivity.this, "200", Toast.LENGTH_SHORT).show();
                }

                if(response.code() == 401){
                    Toast.makeText(RegisterThreeActivity.this, "401", Toast.LENGTH_SHORT).show();
                }
            }

            @Override
            public void onFailure(Call<User> call, Throwable t) {

            }
        });

接口：

 @FormUrlEncoded
    @POST("api/check")
    Call<User> smsPincodeCheck(@Field("email") String email,  @Field("phoneNumber") String phoneNumber);

用户模型：

public class User implements Serializable {
    private final static String TAG = "User";

    @SerializedName("user")
    @Expose
    private Long id;
    private Long commerceId;
    private String email;
    private String name;
    private String lastname;
    private String dni;
    private String companyCharge;
.
.
. Getters and setters

JSON：

这是我想要发送到服务器的json：

{
  "phoneNumber" : "444444444", 
  "email" : "example@example.com"
}

编辑1：

Call<ResponseSMS> call = apiService.smsPincodeCheck(getPhoneNuber, email);
    call.enqueue(new Callback<ResponseSMS>() {
        @Override
        public void onResponse(Call<ResponseSMS> call, Response<ResponseSMS> response) {
            if(response.isSuccess()){
                Toast.makeText(RegisterThreeActivity.this, "200", Toast.LENGTH_SHORT).show();
            }

            if(response.code() == 401){
                Toast.makeText(RegisterThreeActivity.this, "401", Toast.LENGTH_SHORT).show();
            }
        }

        @Override
        public void onFailure(Call<ResponseSMS> call, Throwable t) {
            Toast.makeText(RegisterThreeActivity.this, "Failure", Toast.LENGTH_SHORT).show();
        }
    });
}

接口：

Call<ResponseSMS> smsPincodeCheck(@Field("email") String email, @Field("phoneNumber") String phoneNumber);

型号：

public class ResponseSMS {
    public String response;
    public int status;

    public String getResponse() {
        return response;
    }

    public void setResponse(String response) {
        this.response = response;
    }

    public int getStatus() {
        return status;
    }

    public void setStatus(int status) {
        this.status = status;
    }
}

Answer 1

现在，您需要创建一个单独的模型类来处理Web服务的响应：

public class Response { public String response; public int status; }

并在改装的网络服务电话中将此课程替换为User。

将User替换为Response： 1. smsPincodeCheck ApiInterface接口的Call<Response> call = apiService.smsPincodeCheck(getPhoneNuber, email);方法。 2. call.enqueue(new Callback<Response>() { @Override public void onResponse(Call<Response> call, Response<Response> response) { if(response.isSuccess()){ Toast.makeText(RegisterThreeActivity.this, "200", Toast.LENGTH_SHORT).show(); } if(response.code() == 401){ Toast.makeText(RegisterThreeActivity.this, "401", Toast.LENGTH_SHORT).show(); } } @Override public void onFailure(Call<Response> call, Throwable t) { } }); 3.

<link href="https://fonts.googleapis.com/icon?family=Material+Icons" rel="stylesheet" type="text/css">

<div class="container">
  <ul class="list-inline">
    <li><a href="#"><span class="material-icons">&#xE858;</span></a></li>
    <li><a href="#"><span class="material-icons">&#xE857;</span></a></li>    
  </ul>
</div>

具有改进功能的Post方法永远不会在OnResponse或OnFailure中触发

1 个答案: