Question

我正在尝试使用改造上传文件，将其发送到服务器端并将该文件保存在我的上传文件夹中。

这是我的改造API实例：

@Multipart
@POST("file/uploaddocument")
Call<ResponseBody> uploadFile(@Part MultipartBody.Part file );

UploadFile：

private void uploadFile(Uri fileUri) {
    // create upload service client
    Retrofit retrofit = new Retrofit.Builder()
            .baseUrl(BASE_URL)
            .addConverterFactory(GsonConverterFactory.create())
            .build();

    MyApiEndpointInterface apiService =
            retrofit.create(MyApiEndpointInterface.class);

    // https://github.com/iPaulPro/aFileChooser/blob/master/aFileChooser/src/com/ipaulpro/afilechooser/utils/FileUtils.java
    // use the FileUtils to get the actual file by uri
    File file = Utils.getFileForUri(fileUri);
    // create RequestBody instance from file
    RequestBody requestFile =
            RequestBody.create(
                    MediaType.parse(getContentResolver().getType(fileUri)),
                    file
            );

    // MultipartBody.Part is used to send also the actual file name
    MultipartBody.Part body =
            MultipartBody.Part.createFormData("file", file.getName(), requestFile);

    // finally, execute the request
    Call<ResponseBody> call = apiService.uploadFile(body);
    call.enqueue(new Callback<ResponseBody>() {
        @Override
        public void onResponse(Call<ResponseBody> call,
                               Response<ResponseBody> response) {
            UploadProgressDialog.dismiss();
            Log.v("Upload", "success");
        }

        @Override
        public void onFailure(Call<ResponseBody> call, Throwable t) {
            UploadProgressDialog.dismiss();
            Log.e("Upload error:", t.getMessage());
        }
    });
}

我在点击按钮时调用此方法：

uploadFile(myfileuri);

这是我的Web API调用（这是否正确？如果不是如何从客户端接受图像？）

 <HttpPost>
    <Route("api/File/UploadDocument", Name:="UploadDocument")>
    Public Function Upload() As HttpResponseMessage
        Try
            Dim UploadedPath As String = HttpContext.Current.Server.MapPath("~/UploadedFiles")
            Dim httpRequest = HttpContext.Current.Request
            If httpRequest.Files.Count > 0 Then
                For Each file As String In httpRequest.Files

                    Dim postedFile = httpRequest.Files(file)
                    postedFile.SaveAs(UploadedPath + "/")
                Next
            Else

            End If
            Dim message = Request.CreateResponse(HttpStatusCode.OK, "True")
            Return message
        Catch ex As Exception
            Return Request.CreateErrorResponse(HttpStatusCode.BadRequest, ex)
        End Try
    End Function

Answer 1

我就是这样做的......在这里张贴以便其他人可能会发现它很有用

<HttpPost, Route("api/UploadFile")>
    Public Function Post() As HttpResponseMessage
        Try
            Dim httpRequest = HttpContext.Current.Request
            If httpRequest.Files.Count < 1 Then
                Return Request.CreateResponse(HttpStatusCode.BadRequest)
            End If

            For Each file As String In httpRequest.Files
                Dim postedFile = httpRequest.Files(file)
                Dim filePath = HttpContext.Current.Server.MapPath("~/UploadedFiles/" + postedFile.FileName)
                ' NOTE: To store in memory use postedFile.InputStream
                postedFile.SaveAs(filePath)
            Next
            Return Request.CreateResponse(HttpStatusCode.NoContent)
        Catch ex As Exception
            Return Request.CreateErrorResponse(HttpStatusCode.BadRequest, ex)
        End Try
    End Function

相应的web api函数接受文件

1 个答案: