java c#HttpClient将文件上载到spring rest服务

Question

问题： 我有一个Java spring rest服务来上传文件（大尺寸）。 我想用一把。NET httpClient（或其他.NET客户端）调用上载服务

问题:

1. 似乎发送大文件的最佳选择是多部分文件，那么互操作性是什么呢
2. 如果不可能，最好的选择是什么

谢谢!

Answer 1

答案如下： 我可以将带有多部分附件的文件从c#客户端发送到Java JAX Rest Webservice

 try
        {
            using (
            var client = new HttpClient())
            using (var form = new MultipartFormDataContent())
            {
                using (var stream = new FileStream(fileName, FileMode.Open, FileAccess.Read, FileShare.ReadWrite)) {
                    using (var fileContent = new StreamContent(stream)) {

                        fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment") {FileName = fileName, DispositionType = DispositionTypeNames.Attachment, Name = "fileData"};

                        form.Add(fileContent);
                        // only for test purposes, for stable environment, use ApiRequest class.
                        response = client.PostAsync(url, form).Result;
                    }
                }
            }

            return response.RequestMessage != null ? response.ReasonPhrase : null;
        }
        catch (Exception ex)
        {
            TraceManager.TraceError("Post Asyn Request to " + url + " \n" + ex.Message, ex);
            throw;
        }

Answer 2

HTTP是一个独立于操作系统平台和编程语言的标准，因此在您的情况下，互操作性不应该有任何问题。net客户端符合这些标准

Answer 3

java spring boot

@RequestMapping(value="/upload", method=RequestMethod.POST)
public @ResponseBody String upload(@RequestParam("FileParam") MultipartFile file){
    InputStream fromClient=file.getInputStream();
    ...do stuff with the database/ process the input file...

#

HttpClient client = new HttpClient();
MultipartFormDataContent form = new MultipartFormDataContent();
FileInfo file = new FileInfo(@"<file path>");
form.Add(new StreamContent(file.OpenRead()),"FileParam",file.Name);
HttpResponseMessage response = await client.PostAsync("http://<host>:<port>/upload", form);
Console.WriteLine(response.StatusCode);
Console.WriteLine(response.ReasonPhrase);
Console.WriteLine(response.ToString());
Console.WriteLine(Encoding.ASCII.GetString(await response.Content.ReadAsByteArrayAsync()));