C# .NET 中的异步 POST 请求

Question

我正在尝试使用 HttpClient / HttpContent 通过网络上传数据 但是我似乎找不到以这种方式发送文件的正确方法。

这是我当前的代码：

    private async Task<APIResponse> MakePostRequest(string RequestUrl, string Content)
    {
        HttpClient  httpClient = new HttpClient();
        HttpContent httpContent = new StringContent(Content);
        APIResponse serverReply = new APIResponse();

        httpContent.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
        try {
            Console.WriteLine("Sending Request: " + RequestUrl + Content);
            HttpResponseMessage response = await httpClient.PostAsync(RequestUrl, httpContent).ConfigureAwait(false);
        }
        catch (HttpRequestException hre)
        {
            Console.WriteLine("hre.Message");
        }
        return (serverReply);
    }

内容是这种形式的字符串：paramname=value&param2name=value&param3name=value.. 关键是我必须通过这个请求实际发送一个文件（照片）。

似乎每个参数都可以正常工作，但文件本身（我必须​​在发布请求中发送两个身份验证密钥，并且它们被识别）

我以这种方式将图片作为字符串检索，这可能是它失败的主要原因之一？ ：/

    byte[] PictureData = File.ReadAllBytes(good_path);
    string encoded = Convert.ToBase64String(PictureData);

我做错了吗？是否有另一种更好的方法来创建正确的 POST 请求（它必须是异步的并支持文件上传）

谢谢。

Answer 1

主要问题可能来自我混合字符串和文件数据的事实。

这解决了它：

    public async Task<bool> CreateNewData (Data nData)
    {
        APIResponse serverReply;

        MultipartFormDataContent form = new MultipartFormDataContent ();

        form.Add (new StringContent (_partnerKey), "partnerKey");
        form.Add (new StringContent (UserData.Instance.key), "key");
        form.Add (new StringContent (nData.ToJson ()), "erList");

        if (nData._FileLocation != null) {
            string good_path = nData._FileLocation.Substring (7); // Dangerous
            byte[] PictureData = File.ReadAllBytes (good_path);
            HttpContent content = new ByteArrayContent (PictureData);
            content.Headers.Add ("Content-Type", "image/jpeg");
            form.Add (content, "picture_0", "picture_0");
        }

        form.Add (new StringContent (((int)(DateTime.Now.ToUniversalTime () -
            new DateTime (1970, 1, 1, 0, 0, 0, DateTimeKind.Utc)).TotalSeconds).ToString ()), "time");
        serverReply = await MakePostRequest (_baseURL + "Data-report/create", form);

        if (serverReply.Status == SERVER_OK)
            return (true);
        Android.Util.Log.Error ("MyApplication", serverReply.Response);
        return (false);
    }

    private async Task<APIResponse> MakePostRequest (string RequestUrl, MultipartFormDataContent Content)
    {
        HttpClient httpClient = new HttpClient ();
        APIResponse serverReply = new APIResponse ();

        try {
            Console.WriteLine ("MyApplication - Sending Request");
            Android.Util.Log.Info ("MyApplication", " Sending Request");
            HttpResponseMessage response = await httpClient.PostAsync (RequestUrl, Content).ConfigureAwait (false);
            serverReply.Status = (int)response.StatusCode;
            serverReply.Response = await response.Content.ReadAsStringAsync ();
        } catch (HttpRequestException hre) {
            Android.Util.Log.Error ("MyApplication", hre.Message);
        }
        return (serverReply);
    }

主要使用Multipart Content，设置Content Type并通过字节数组似乎已经完成了这项工作。

Answer 2

您需要将数据作为 base64 编码的字符串发送吗？如果您要发送任意字节（即照片），则可以发送未编码的字节，如果您使用 ByteArrayContent 类：

private async Task<APIResponse> MakePostRequest(string RequestUrl, byte[] Content)
{
    HttpClient  httpClient = new HttpClient();
    HttpContent httpContent = new ByteArrayContent(Content);
    APIResponse serverReply = new APIResponse();

    try {
        Console.WriteLine("Sending Request: " + RequestUrl + Content);
        HttpResponseMessage response = await httpClient.PostAsync(RequestUrl, httpContent).ConfigureAwait(false);
        // do something with the response, like setting properties on serverReply?
    }
    catch (HttpRequestException hre)
    {
        Console.WriteLine("hre.Message");
    }

    return (serverReply);
}