使用 PostAsync 从 HTML 表单上传文件

Question

我只想使用 WebAPI 将文件上传到服务器。我想在我的 MVC HomeController 中调用 WebApi。当我直接使用 WebAPI 时，它可以完美地上传文件，但是当我想使用 PostAsync 在我的控制器中调用 WebAPI 时，由于某种原因它不会将文件发送到 api。

这是我的 MVC 控制器中的操作：

 public async Task<ActionResult> Upload(HttpPostedFileBase upload)
    {
        ViewBag.Title = "Upload Files";
        if (upload != null && upload.ContentLength > 0)
        {

            var client = new HttpClient();

            string FullUrl = Request.Url.Scheme + System.Uri.SchemeDelimiter + Request.Url.Host + ":" + Request.Url.Port + "/api/values";
            StringContent httpcontent = new StringContent(upload.FileName);

            var response = await client.PostAsync(FullUrl, httpcontent);

            var result = await response.Content.ReadAsAsync<HttpResponseMessage>();

        }


        return View();
    }

这是 .cshtml 文件：

<form name="form1" method="post" action="/Home/Upload" enctype="multipart/form-data">

<div>
    <label for="upload">Choose a File</label>
    <input name="upload" type="file" />
</div>
<div>
    <input type="submit" value="Submit" />
</div>

最后这是 WebApi 控制器：

 [HttpPost]
    public HttpResponseMessage PostFile()
    {
        HttpResponseMessage result = null;
        var httpRequest = HttpContext.Current.Request;

        if (httpRequest.Files.Count > 0)
        {
            foreach (string file in httpRequest.Files)
            {
                var postedFile = httpRequest.Files[file];
                var filePath = HttpContext.Current.Server.MapPath("~/Files/" + postedFile.FileName);
                postedFile.SaveAs(filePath);
            }
            result = Request.CreateResponse(HttpStatusCode.Created);
        }
        else
        {
            result = Request.CreateResponse(HttpStatusCode.BadRequest);
        }

        return result;
    }

Answer 1

当表单内容post时，它设置multipart/form-data请求的内容类型。 您可以尝试将请求内容类型明确设置为 multipart/form-data。

var requestContent = new MultipartFormDataContent();
var fileContent = new StreamContent(upload.InputStream);
fileContent.Headers.ContentType = upload.ContentType;
requestContent.Add(fileContent, upload.FileName, upload.FileName);

client.PostAsync(url, requestContent);