Question

我正在尝试将文件发布到PushBullet API服务。

API的信息网站显示了一个示例：

curl -i https://api.pushbullet.com/api/pushes \
-u API_KEY:
-F device_iden=u1qSJddxeKwOGuGW
-F type=file
-F file=@test.txt

我正在尝试使用C＃

中的HttpClient

我知道如何使用Httpclient使用MultipartFormDataContent发布文件，但是如何将device_iden和类型信息添加到客户端？

我正在使用这个

using (var content = new MultipartFormDataContent())
            {
                try
                {
                    content.Add(new StreamContent(new FileStream(pathFile, FileMode.Open, FileAccess.Read)));
                    var resp = wc.PostAsync(new Uri(baseUri, "api/pushes"), content);

                }
                catch (Exception)
                {

                    throw;
                }
            }

[[新增]]

使用cURL和fiddler这将是POST

POST http://\/ HTTP/1.1
Authorization: Basic d45YQkxueWswSmxQRTYFjc1deUzNo8UtueVZpaktIZm34anVqdU9NZerWYmFlOp==
User-Agent: curl/7.33.0
Host: \
Accept: */*
Proxy-Connection: Keep-Alive
Content-Length: 787
Expect: 100-continue
Content-Type: multipart/form-data; boundary=------------------------886f3981539a91b3

--------------------------886f3981539a91b3
Content-Disposition: form-data; name="device_iden"

ujuOMfjVbaedjz7O3P0Jl6
--------------------------886f3981539a91b3
Content-Disposition: form-data; name="type"

file
--------------------------886f3981539a91b3
Content-Disposition: form-data; name="file"; filename="img.gif"
Content-Type: application/octet-stream
GIF89a������@�;A�;A�;A�;B�<B�<������!�Created with GIMP�!�
��,����������x�����I��8KȻ�`(�$��h��l{�p�tm߮��x����P,���҃l:q���I���؅u�Mf����
��d�y}LG��{���J�>���W};������^�1�����������

--------------------------886f3981539a91b3--

Answer 1

在与fiddler一起尝试后，终于可以发帖了：

首先：

public static void PushFile(string pathFile, string iddev)
        {         
            string name = Path.GetFileName(pathFile);
            using (var wc = GetClient())
            {
                using (var multiPartCont = new MultipartFormDataContent())
                {
                    multiPartCont.Add(addStringContent("device_iden", iddev));
                    multiPartCont.Add(addStringContent("type","file"));
                    multiPartCont.Add(addStreamContent(new FileStream(pathFile,FileMode.Open),name ));

                    try
                    {
                        var resp = wc.PostAsync(new Uri(baseUri, "api/pushes"), multiPartCont);
                        Task<string> result = resp.Result.Content.ReadAsStringAsync();
                        //string resultado = result.Result;
                    }
                    catch (Exception)
                    {
                        throw;
                    }
                }
            }            
        }

然后创建字符串内容和流内容的方法

private static StreamContent addStreamContent(Stream stream, string filename )
    {
        var fileContent = new StreamContent(stream);
        fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
        {
            Name = "\"file\"",
            FileName = "\""+filename+"\""
        };
        fileContent.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
        return fileContent;
    }
private static StringContent addStringContent(string name, string content)
{
    var fileContent = new StringContent(content);
    fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
    {
        Name = "\"" + name + "\""
    };
    return fileContent;
}

Answer 2

curl中的-u是http身份验证。看起来pushbullet希望您使用API密钥登录。使用httpclient，你会使用;

httpClient.DefaultRequestHeaders.Authorization = CreateBasicHeader("username", "password");

其余的将是GET或POST变量（我不确定pushbullet的预期）。在您的代码中，您可以通过POST设置它。因此，您可以使用;

附加到“content”变量

 "device_iden=u1qSJddxeKwOGuGW&type=file";

Answer 3

创建-u数据的另一种方法是使用HttpClientHandler

创建NetworkCredential

var handler = new HttpClientHandler
{
    Credentials = new NetworkCredential(API_KEY, "")
};

var client = new HttpClient(handler);

这将创建一个基本的http身份验证标头

cURL发布到HTTPCLient C＃

3 个答案: