在 Retrofit 2 中上传文件答案

【问题标题】：Upload file in Retrofit 2在 Retrofit 2 中上传文件
【发布时间】：2016-08-31 04:01:12
【问题描述】：

我尝试了以下操作，但在响应时我收到 500 错误（内部服务器错误）--帮助我为上述屏幕截图中的请求设计接口...谢谢

@Multipart
@POST("myrecord")
Call<ResponseBody> addRecord(@Query("token") String token,@Query("userid") int userId,
                             @Query("name") String name, @Part("file") RequestBody file);


File file = new File(getRealPathFromURI(data.getData()));
RequestBody requestFile = RequestBody.create(MediaType.parse("image/*"), getRealPathFromURI(data.getData()));` 
Call<ResponseBody> responseBodyCall = service.addRecord(token, userId,
                "newFileName", requestFile);
        responseBodyCall.enqueue(new Callback<ResponseBody>() {
            @Override
            public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                Log.d("Response", "="+response.code());
                Log.d("Response", "= "+response.message());

            }

            @Override
            public void onFailure(Call<ResponseBody> call, Throwable t) {
                Log.d("failure", "message = " + t.getMessage());
                Log.d("failure", "cause = " + t.getCause());
            }
        });`

【问题讨论】：

尝试删除@Multipart
@MobDev 没用兄弟..

标签： android android-studio retrofit2

【解决方案1】：

以下代码有效:)

 @Multipart
@POST("myrecord")
Call<ResponseBody> addRecord(@Query("token") String token, @Query("userid") int userId,
                             @Query("name") String name, @Part MultipartBody.Part file);



 @Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    if ((requestCode == FILE_SELECT_CODE) && (resultCode == -1)) {

        File file = new File(getRealPathFromURI(data.getData()));

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

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

        Call<ResponseBody> responseBodyCall = service.addRecord(token, userId, "fileName", multipartBody);
        responseBodyCall.enqueue(new Callback<ResponseBody>() {
            @Override
            public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                Log.d("Success", "success "+response.code());
                Log.d("Success", "success "+response.message());

            }

            @Override
            public void onFailure(Call<ResponseBody> call, Throwable t) {
                Log.d("failure", "message = " + t.getMessage());
                Log.d("failure", "cause = " + t.getCause());
            }
        });

    }
}

【讨论】：

请您更新getRealPathFromURI()方法？
我有一个问题，我正在尝试这段代码，但现在我很困惑如何创建在上述 android 代码中正常工作的 asp.net Web 服务。我已经试过了。有关更多信息，请参阅此stackoverflow.com/q/57909118/10325004 并请帮助我。

【解决方案2】：

@Multipart
@POST("myrecord")
Call<ResponseBody> addRecord(@Part("file") RequestBody file,@Part MultipartBody.Part file,
@Query("token") String token,@Query("userid") int userId,@Query("name") String name);

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

MultipartBody.Part body =
        MultipartBody.Part.createFormData("picture", file.getName(), requestFile);


String descriptionString = "your description";

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

有关更多信息，请查看此链接： https://futurestud.io/blog/retrofit-2-how-to-upload-files-to-server

【讨论】：

【解决方案3】：

如果您想在正文中以二进制形式发送文件而不使用 multipart，您可以从代码中删除 @Multipart 注释并使用 @Body 注释。好像

@POST("myrecord")
Call<ResponseBody> addRecord(@Query("token") String token,@Query("userid") int userId,
                             @Query("name") String name, @Body RequestBody file);


File file = new File(getRealPathFromURI(data.getData()));
RequestBody requestFile = RequestBody.create(MediaType.parse("image/*"), getRealPathFromURI(data.getData()));
Call<ResponseBody> responseBodyCall = service.addRecord(token, userId,
                "newFileName", requestFile);
        responseBodyCall.enqueue(new Callback<ResponseBody>() {
            @Override
            public void onResponse(Call<ResponseBody> call, Response<ResponseBody> response) {
                Log.d("Response", "="+response.code());
                Log.d("Response", "= "+response.message());
        }

        @Override
        public void onFailure(Call<ResponseBody> call, Throwable t) {
            Log.d("failure", "message = " + t.getMessage());
            Log.d("failure", "cause = " + t.getCause());
        }
    });`

【讨论】：

这种方式和Multipart方式有什么区别？在这两种情况下，文件都将作为二进制发送，对吗？
@Mena multipart 用于发送“多个主体”，例如多个文件或文件和 json，或多个 json（如果需要）。如果您只需要发送 1 个文件，则此方法更好。无论如何取决于服务器端配置

【解决方案4】：

    public String getRealPathFromURI(Context context, Uri contentUri) {
    Log.d("imin", "onClick: in image conversion");

    Cursor cursor = null;
    try {
        String[] proj = {MediaStore.Images.Media.DATA};
        cursor = context.getContentResolver().query(contentUri, proj, null, null, null);
        int column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
        cursor.moveToFirst();
        Log.d("imin", "onClick: in image conversion try");

        return cursor.getString(column_index);
    } finally {
        Log.d("imin", "onClick: in image conversion finally");

        if (cursor != null) {
            cursor.close();
        }
    }
}

【讨论】：