如何通过 HttpClient PostAsync 方法将文件和参数上传到远程服务器？

Question

我正在尝试将文件从我的桌面应用程序上传到远程服务器。 在浏览了一段时间之后，这种方法似乎是最干净的方法。 问题是在服务器端都没有收到参数。 我错过了什么？

    private void AddFile(FileInfo fileInfo, int folderId)
    {
        using (var handler = new HttpClientHandler() {CookieContainer = _cookies})
        {
            using (var client = new HttpClient(handler) {BaseAddress = new Uri(_host)})
            {
                var requestContent = new MultipartFormDataContent();
                var fileContent = new StreamContent(fileInfo.Open(FileMode.Open));
                var folderContent = new StringContent(folderId.ToString(CultureInfo.InvariantCulture));
                requestContent.Add(fileContent, "file", "file");
                requestContent.Add(folderContent, "folderId", "folderId");

                client.PostAsync("/Company/AddFile", requestContent);
            }
        }
    }

编辑：这是服务器端期望的签名：

    [HttpPost]
    public ActionResult AddFile(HttpPostedFileBase file, int folderId)

Answer 1

经过大量的反复试验，我得到了它。 有一些问题。 1）引号中的参数名称预期2）我遗漏了一堆标题信息。 这是工作代码。

    private void AddFile(FileInfo fileInfo, int folderId)
    {
        using (var handler = new HttpClientHandler() {CookieContainer = _cookies})
        {
            using (var client = new HttpClient(handler) {BaseAddress = new Uri(_host)})
            {
                var requestContent = new MultipartFormDataContent();
                var fileContent = new StreamContent(fileInfo.OpenRead());
                fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
                    {
                        Name = "\"file\"",
                        FileName = "\"" + fileInfo.Name + "\""
                    };
                fileContent.Headers.ContentType =
                    MediaTypeHeaderValue.Parse(MimeMapping.GetMimeMapping(fileInfo.Name));
                var folderContent = new StringContent(folderId.ToString(CultureInfo.InvariantCulture));

                requestContent.Add(fileContent);
                requestContent.Add(folderContent, "\"folderId\"");

                var result = client.PostAsync("Company/AddFile", requestContent).Result;
            }
        }

Answer 2

在客户端： using（var formData = new MultipartFormDataContent（））{formData.Add（new StreamContent（new MemoryStream（bytes）），“file”，Model.BaseFileName）; string result = client.PostAsync（“ https://fakeurl.com/Controller/Action?inputParams= ”+ Params，formData）.Result.Content.ReadAsStringAsync（）。Result;}

在服务器端： var httpRequest = System.Web.HttpContext.Current.Request;

            if (httpRequest.Files.Count > 0)
            {
                var docfiles = new List<string>();
                //foreach (string file in httpRequest.Files)
                //{
                HttpPostedFile postedFile = httpRequest.Files[0];
                // Initialize the stream.
                Stream mstream = postedFile.InputStream;

                byte[] byteArray = new byte[postedFile.ContentLength];
                postedFile.InputStream.Read(byteArray, 0, postedFile.ContentLength);

Answer 3

本文提出一个完整的解决方案：https ://makolyte.com/csharp-how-to-send-a-file-with-httpclient/

上传单个文件的代码：

var filePath = @"C:\house.png";

using (var multipartFormContent = new MultipartFormDataContent())
{
    //Load the file and set the file's Content-Type header
    var fileStreamContent = new StreamContent(File.OpenRead(filePath));
    fileStreamContent.Headers.ContentType = new MediaTypeHeaderValue("image/png");

    //Add the file
    multipartFormContent.Add(fileStreamContent, name: "file", fileName: "house.png");

    //Send it
    var response = await httpClient.PostAsync("https://localhost:12345/files/", multipartFormContent);
    response.EnsureSuccessStatusCode();
    return await response.Content.ReadAsStringAsync();
}

上传文件和一些表单字段的代码：

var filePath = @"C:\house.png";

using (var multipartFormContent = new MultipartFormDataContent())
{
    //Add other fields
    multipartFormContent.Add(new StringContent("123"), name: "UserId");
    multipartFormContent.Add(new StringContent("Home insurance"), name: "Title");

    //Add the file
    var fileStreamContent = new StreamContent(File.OpenRead(filePath));
    fileStreamContent.Headers.ContentType = new MediaTypeHeaderValue("image/png");
    multipartFormContent.Add(fileStreamContent, name: "file", fileName: "house.png");

    //Send it
    var response = await httpClient.PostAsync("https://localhost:12345/files/", multipartFormContent);
    response.EnsureSuccessStatusCode();
    return await response.Content.ReadAsStringAsync();
}