修改2解析传出数据

Question

你好StackOverflow的伙计们！这是我的第一个帖子，真的是我的大问题与改造2，我希望一些人帮助解决。

这是我的接口类：

public interface PicUploadInterface {
@Multipart
@POST("user")
Call<UserModelResponse> uploadMultipleFiles(@Part("Authorization")  RequestBody authorization,
                          @Part("description") RequestBody description,
                          @Part MultipartBody.Part filePart);

和apimanager类：

public class Apimanager {

public static void upload(String coda, File file) {
    // create upload service client
    final String API_BASE_URL = "http://10.0.3.2/web2Demo/public/";
    OkHttpClient.Builder httpClient = new OkHttpClient.Builder();

    HttpLoggingInterceptor logging = new HttpLoggingInterceptor();
    // set your desired log level
    logging.setLevel(HttpLoggingInterceptor.Level.BODY);
    httpClient.addInterceptor(logging);  

    Retrofit builder =
            new Retrofit.Builder()
                    .baseUrl(API_BASE_URL)
                    .addConverterFactory(GsonConverterFactory.create())
                    .client(httpClient.build())
                    .build();

    PicUploadInterface service = builder.create(PicUploadInterface.class);
    RequestBody requestFile = RequestBody.create(MediaType.parse("multipart/form-data"), file);

    MultipartBody.Part filePart  = MultipartBody.Part.createFormData("uploaded_file", file.getName(), requestFile);

    String descriptionString = coda;
    String AuthString = "Send picture";
    RequestBody description = RequestBody.create(MediaType.parse("text/plain"), descriptionString);
    RequestBody authorization = RequestBody.create(MediaType.parse("text/plain"), AuthString);

    Call<UserModelResponse> call = service.uploadMultipleFiles(authorization, description, filePart);
    call.enqueue(new Callback<UserModelResponse>() {
        @Override
        public void onResponse(Call<UserModelResponse> call,
                               Response<UserModelResponse> response) {
            Log.v("Upload", "success");
        }
        @Override
        public void onFailure(Call<UserModelResponse> call, Throwable t) {
            Log.e("Upload error:", t.getMessage());
        }
    });
}

我以这种方式使用代码：

      File file = new File(textViewFile.getText().toString()); // getting this string "/storage/emulated/0/bichooser/1476172212892.jpg"
      connectivity.com.pack.Apimanager.upload(name, file);

最后，我的HttpLoggingInterceptor显示，我在restApi中找不到什么和如何解析它

我找不到从restApi改版2.1.0到restApi的内容，以及哪一部分是我的字符串或图像数据？我确实根据stackoverflow提示和另一个站点改变了很多次，但我的结果几乎是一样的！（我做了短图像二进制从拦截器日志传出）如果有人有想法，请分享。谢谢

Answer 1

您是否尝试在服务呼叫中直接使用：

RequestBody requestFile = RequestBody.create(MediaType.parse("multipart/form-data"), file); 

...

Call<UserModelResponse> call = service.uploadMultipleFiles(authorization, description, requestFile);

和更新：

@Multipart
@POST("user")
Call<UserModelResponse> uploadMultipleFiles(@Part("Authorization")  RequestBody authorization,
                          @Part("description") RequestBody description,
                          @Part("file\"; filename=\"myFilename\" ") RequestBody filePart);

另外，如果您想查看对restApi的调用，您可以使用Charles代理作为反向代理，查看您的调用和响应。