Question

我正在建立一个新的android项目并使用改造，我的改造功能可以在模拟器（NOX）和邮递员中正常工作，但是当我尝试在移动设备上构建我的应用程序时，改造总是会陷入onFailure，谁能给我解决方案？我的API在公共托管上发布，

这就是我所说的改造

 private APIInterface getInterfaceService() {
        Retrofit retrofit = new Retrofit.Builder()
                .baseUrl(BASE_URL)
                .addConverterFactory(GsonConverterFactory.create())
                .build();
        final APIInterface mInterfaceService = retrofit.create(APIInterface.class);
        return mInterfaceService;
    }

 private void loginInterface(final String username, final String password){
        APIInterface mApiService = this.getInterfaceService();
        Call<Response> mService = mApiService.loginRequest(username,password);
        mService.enqueue(new Callback<Response>() {
            @Override
            public void onResponse(Call<Response> call, retrofit2.Response<Response> response) {
                if(response.body().getValue()==1){
                    Toast.makeText(Login.this,"Welcome",Toast.LENGTH_SHORT).show();
                    Intent intent = new Intent(getApplicationContext(),HomePage.class);
                    startActivity(intent);
                    finish();
                }else{
                    Toast.makeText(Login.this,"Invalid Username or Password",Toast.LENGTH_SHORT).show();
                }
            }

            @Override
            public void onFailure(Call<Response> call, Throwable t) {
                Toast.makeText(Login.this,t.getMessage(),Toast.LENGTH_SHORT).show();
            }
        });
    }

我的回应

 public class Response {

    @SerializedName("value")
    @Expose
    private Integer value;
    @SerializedName("result")
    @Expose
    private List<User> result = null;

    public Integer getValue() {
        return value;
    }

    public void setValue(Integer value) {
        this.value = value;
    }

    public List<User> getResult() {
        return result;
    }

    public void setResult(List<User> result) {
        this.result = result;
    }

}

用户模型

 public class User {
    @SerializedName("id")
    @Expose
    private String id;
    @SerializedName("username")
    @Expose
    private String username;
    @SerializedName("password")
    @Expose
    private String password;
    @SerializedName("email")
    @Expose
    private String email;
    @SerializedName("image")
    @Expose
    private Object image;
    @SerializedName("point")
    @Expose
    private String point;
    @SerializedName("reputation")
    @Expose
    private String reputation;
    @SerializedName("role")
    @Expose
    private String role;

    public User(String id, String username, String password, String email, Object image, String point, String reputation, String role) {
        this.id = id;
        this.username = username;
        this.password = password;
        this.email = email;
        this.image = image;
        this.point = point;
        this.reputation = reputation;
        this.role = role;
    }

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public Object getImage() {
        return image;
    }

    public void setImage(Object image) {
        this.image = image;
    }

    public String getPoint() {
        return point;
    }

    public void setPoint(String point) {
        this.point = point;
    }

    public String getReputation() {
        return reputation;
    }

    public void setReputation(String reputation) {
        this.reputation = reputation;
    }

    public String getRole() {
        return role;
    }

    public void setRole(String role) {
        this.role = role;
    }
}

API接口

public interface APIInterface {
    @FormUrlEncoded
    @POST("login.php")
    Call<Response> loginRequest(@Field("username") String username,
                            @Field("password") String password);
}

我从t.message收到此消息 “网络安全政策不允许与{my api url}进行CLEARTEXT通信”

将其添加到清单后

android:usesCleartextTraffic="true"

我有了这个新的 'java.lang.IllegalStateException：应为BEGIN_OBJECT，但在第1行第1列的路径$ STRING处

我的JSON响应就是这样

{"value":1,"result":[{"id":"1","username":"username","password":"password","email":"email","image":null,"point":"0","reputation":"0","role":"2"}]}

这是我的邮递员回复

Answer 1

使用this网站在Java中生成正确的响应类

您的响应类应为：

public class Response implements Serializable
{

@SerializedName("value")
@Expose
private long value;
@SerializedName("result")
@Expose
private List<Result> result = null;
private final static long serialVersionUID = -7121130042760098410L;

public long getValue() {
return value;
}

public void setValue(long value) {
this.value = value;
}

public List<Result> getResult() {
return result;
}

public void setResult(List<Result> result) {
this.result = result;
}

}
-----------------------------------com.example.Result.java-----------------------------------

package com.example;

import java.io.Serializable;
import com.google.gson.annotations.Expose;
import com.google.gson.annotations.SerializedName;

public class Result implements Serializable
{

@SerializedName("id")
@Expose
private String id;
@SerializedName("username")
@Expose
private String username;
@SerializedName("password")
@Expose
private String password;
@SerializedName("email")
@Expose
private String email;
@SerializedName("image")
@Expose
private Object image;
@SerializedName("point")
@Expose
private String point;
@SerializedName("reputation")
@Expose
private String reputation;
@SerializedName("role")
@Expose
private String role;
private final static long serialVersionUID = 7267197789545166983L;

public String getId() {
return id;
}

public void setId(String id) {
this.id = id;
}

public String getUsername() {
return username;
}

public void setUsername(String username) {
this.username = username;
}

public String getPassword() {
return password;
}

public void setPassword(String password) {
this.password = password;
}

public String getEmail() {
return email;
}

public void setEmail(String email) {
this.email = email;
}

public Object getImage() {
return image;
}

public void setImage(Object image) {
this.image = image;
}

public String getPoint() {
return point;
}

public void setPoint(String point) {
this.point = point;
}

public String getReputation() {
return reputation;
}

public void setReputation(String reputation) {
this.reputation = reputation;
}

public String getRole() {
return role;
}

public void setRole(String role) {
this.role = role;
}

}

Answer 2

在项目中尝试使用翻新库时应遵循的事项。

使用此Website将您的回复转换为POJO。然后，需要示例找出这个Answer。

您的BASE URL应该以'/'结尾。

  String BASEURL = "http://www.xxxxxxxx.com/"

当实现改造的回调时。您需要通过选中if(response.isSuccessful())来检查您的回复是否给了您200。

 mService.enqueue(new Callback<Response>() {
        @Override
        public void onResponse(Call<Response> call, retrofit2.Response<Response> response) {
        if(response.isSuccessful()){
           if(response.body().getValue()==1){
                Toast.makeText(Login.this,"Welcome",Toast.LENGTH_SHORT).show();
                Intent intent = new Intent(getApplicationContext(),HomePage.class);
                startActivity(intent);
                finish();
            }else{
                Toast.makeText(Login.this,"Invalid Username or Password",Toast.LENGTH_SHORT).show();
            }
          }else   Toast.makeText(Login.this,response.errorBody().string(),Toast.LENGTH_SHORT).show(); // this will tell you why your api doesnt work most of time
        }

        @Override
        public void onFailure(Call<Response> call, Throwable t) {
            Toast.makeText(Login.this,t.getMessage(),Toast.LENGTH_SHORT).show();
        }
    });

正如我所看到的，您正在使用Response类。响应具有许多预定义的对象，其他库也具有响应的名称。因此，您必须检查导入文件以交叉验证您使用的响应对象是您的包名称，而不是其他响应对象。
与在POSTMAN中一样，您包括9个标头。您必须找出所有9个标头改装示例，并将它们相应地添加到API接口中。

注意：-请遵循以下步骤，我想您会自己解决问题的。

按照上述所有说明操作后，您可以随意提问。

Answer 3

您设置了正确的IP地址吗？

如果在邮递员和仿真器中正常运行，则您的代码运行良好。但是，请确保设置正确的终结点（或您的情况下的BASE_URL）

OR

如果您使用的是ProGuard，请确保禁用它只是为了检查没有proguard的情况下一切正常。

翻新总是响应onFailure

3 个答案: