Put方法在Android中不起作用

Question

• 我正在使用演示APi(http://dummy.restapiexample.com/update)
• 我有一个响应类UpdateResponse
• 我有一个Api类和一个APIInterface
• 和对话框而不是主活动
• 在APIInterface中使用PUT方法有一个@path(id)和三个@fields(name、salary、age)

下面是UpdateResponse类代码

public class UpdateResponse {
@SerializedName("id")
@Expose
private int id;
@SerializedName("name")
@Expose
private String name;
@SerializedName("salary")
@Expose
private String salary;
@SerializedName("age")
@Expose
private String age;

public int getId() {
    return id;
}

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

public String getName() {
    return name;
}

public void setName(String employeeName) {
    this.name = employeeName;
}

public String getSalary() {
    return salary;
}

public void setSalary(String employeeSalary) {
    this.salary = employeeSalary;
}

public String getAge() {
    return age;
}

public void setAge(String employeeAge) {
    this.age = employeeAge;
}
}

Api接口代码如下

public interface ApiInterface {


@FormUrlEncoded
@PUT("api/v1/update/{id}")
Call<UpdateResponse> updateUser(@Path("id") int id,
                                  @Field("name") String name,
                                  @Field("salary") String salary,
                                  @Field("age") String age);

}

public class Api {
private static Retrofit retrofit = null;
public static ApiInterface getClient() {


    if (retrofit==null) {
        retrofit = new Retrofit.Builder()
                .baseUrl("http://dummy.restapiexample.com/")
                .addConverterFactory(GsonConverterFactory.create())
                .build();
    }

    ApiInterface api = retrofit.create(ApiInterface.class);
    return api;
}

}

下面是作为主活动代码的对话框

                String nameStr = name.getText().toString();
                String salaryStr = salary.getText().toString();
                String ageStr = age.getText().toString();
                //idd is getting from mainActivity onitemSelect method, which is having the right id value

               Call<UpdateResponse> call= Api.getClient().updateUser(idd,nameStr,salaryStr,ageStr);
               call.enqueue(new Callback<UpdateResponse>() {
                    @Override
                    public void onResponse(Call<UpdateResponse> call, Response<UpdateResponse> response) {
                        Toast.makeText(c.getApplicationContext(),"Updated Name: "+response.body().getName(),Toast.LENGTH_LONG).show();
                        dismiss();
                    }

                    @Override
                    public void onFailure(Call<UpdateResponse> call, Throwable t) {
                        Toast.makeText(c.getApplicationContext(),"Failure",Toast.LENGTH_LONG).show();
                        dismiss();
                    }
                });

响应呼叫显示：

call:executerCallAdapterFactory$executerCallbackCall@5922“，并在可抛出的测试中显示”com.google.gson.stream.malformedJsonException：Unterminated object at line 1 column 27 path$.error“。

Answer 1

您需要发送一个json正文

如下所示更改api方法签名

@Headers({"Content-Type: application/json"})
@PUT("api/v1/update/{id}")
Call<ResponseBody> updateUser(@Path("id") int id, @Body UpdateResponse body);

使id瞬态化

public class UpdateResponse {
    @SerializedName("id")
    @Expose
    private transient int id;
    //..

 UpdateResponse updateResponse = new UpdateResponse();
 updateResponse.setName(name.getText().toString());
 updateResponse.setSalary(salary.getText().toString());
 updateResponse.setAge(age.getText().toString());

 Call<UpdateResponse> call= Api.getClient().updateUser(idd, updateResponse);