c# - c# HttpClient上传文件到spring rest服务

Question

问题：我有一个 Java spring rest 服务来上传文件（大尺寸）。我想使用 .NET httpClient（或其他 .net 客户端）来调用上传服务。

问题：

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

谢谢！

Answer 1

这就是答案：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，因此如果您的 .net 客户端符合标准，则互操作性应该不会有任何问题。

Answer 3

java 弹簧启动

@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...

C＃

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()));