Question

首先，我知道这个问题已经被问过很多遍了，但是我还没有解决。我在应用程序中使用Retrofit，我想在其中向服务器发送3个变量。不会调用onResponse方法，请检查我下面的代码并建议我该代码有什么问题，谢谢。

onResponse方法

 private void loginByServer() {
    String email = _emailText.getText().toString();
    String password = _passwordText.getText().toString();
    String str = "customers";

    Log.e("email______",email);
    Log.e("password______",password);
    Log.e("str______",str);

    APIService service = ApiClient.getClient().create(APIService.class);
    Call<String> userCall = service.userLogIn(email,password,str);
    userCall.enqueue(new Callback<String>() {
        @Override
        public void onResponse(Call<String> call, retrofit2.Response<String> response) {
          Log.e("Inner Part__________","_________________");
            Log.e("Response_________", String.valueOf(response));
            String string = response.body();
            Log.e("Response body_________", string);
            if(response.message().toString().equalsIgnoreCase("ok"));{
              //  progressDoalog.setVisibility(View.GONE);
                Log.e("Successfully","Loged In__________");
            }
        }
        @Override
        public void onFailure(Call<String> call, Throwable t) {
            t.printStackTrace();
            System.out.println("failed to upload the file");
      //      progressDoalog.setVisibility(View.GONE);
            getWindow().clearFlags(WindowManager.LayoutParams.FLAG_NOT_TOUCHABLE);
        }
    });
}

API客户端

public class ApiClient {
public static final String BASE_URL = "http://192.168.2.59/easyshop/";
private static Retrofit retrofit = null;

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

API服务接口

public interface APIService {

@FormUrlEncoded
@POST("login.php")
Call<String> userLogIn(@Field("email") String email,
                    @Field("password") String password,
                    @Field("who") String who);
 }

用于登录用户的php文件

<?php
include("db_config.php");
if($_SERVER['REQUEST_METHOD']=='POST' && isset($_REQUEST['who'])) {
$username = $_REQUEST['email'];
$password = md5($_REQUEST['password']);
$who = $_REQUEST['who'];

if ($who == "customers") {
    $sql = "SELECT COUNT(*) AS count FROM customers WHERE (Username = '".$username."' or Email = '".$username."') and Password = '".$password."'";
 } else if($who == "retailers") {
     $sql = "SELECT COUNT(*) AS count FROM retailers WHERE (Username = '".$username."' or Email = '".$username."') and Password = '".$password."'";
 }
 $result = mysqli_query($con,$sql);
 $check = mysqli_fetch_array($result);

if ($check['count'] > 0) {
    echo "Success";
} else {
     echo "Failed";
   }
  }
?>

Answer 1

缺少的部分是@POST（“ login.php”）上的/

public interface APIService {
@FormUrlEncoded
@POST("/login.php")
Call<String> userLogIn(@Field("email") String email,
                @Field("password") String password,
                @Field("who") String who);
}

使用改造库时，不会调用onResponse方法

1 个答案: