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。

这是我目前的代码：

    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);
    }

内容是该格式的字符串：paramname = value＆amp; param2name = value＆amp; param3name = value .. 重点是我必须通过此请求实际发送文件（照片）。

它似乎适用于每个参数但文件本身（我必须在post请求中发送两个验证密钥，并且它们被识别）

我按照这种方式将图片作为字符串进行检索，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（它必须是异步并支持文件上载）

感谢。

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，设置内容类型并通过字节数组似乎完成了这项工作。

Answer 2

您是否需要将数据作为base64编码的字符串发送？如果您要发送任意字节（即照片），如果您使用ByteArrayContent类，则可以将它们发送为未编码的字段：

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);
}

C＃.NET中的异步POST请求

2 个答案: