通过HTTP POST发送文件用C＃通过HTTP POST发送文件用C＃(Send a file v

我一直在寻找，并围绕该阅读和可能不细什么真正有用的。

我正在写一个小的C＃赢应用程序，允许用户将文件发送到Web服务器，而不是通过FTP，但通过HTTP POST使用。 认为它像一个web形式，但在Windows应用程序中运行。

我一直在使用这样的事情我的HttpWebRequest对象创建

HttpWebRequest req = WebRequest.Create(uri) as HttpWebRequest 

并且还设置Method ， ContentType和ContentLength属性。 但是那遥远的我可以走了。

这是我的代码：

HttpWebRequest req = WebRequest.Create(uri) as HttpWebRequest;
req.KeepAlive = false;
req.Method = "POST";
req.Credentials = new NetworkCredential(user.UserName, user.UserPassword);
req.PreAuthenticate = true;
req.ContentType = file.ContentType;
req.ContentLength = file.Length;
HttpWebResponse response = null;

try
{
    response = req.GetResponse() as HttpWebResponse;
}
catch (Exception e) 
{
}

所以我的问题基本上是我怎么能与C＃通过HTTP POST发送FIE（文本文件，图像，音频等）。

谢谢！

Answer 1:

使用.NET 4.5（或通过添加.NET 4.0 Microsoft.Net.Http从NuGet包）有模拟形式的请求更简单的方法。 下面是一个例子：

private async Task<System.IO.Stream> Upload(string actionUrl, string paramString, Stream paramFileStream, byte [] paramFileBytes)
{
    HttpContent stringContent = new StringContent(paramString);
    HttpContent fileStreamContent = new StreamContent(paramFileStream);
    HttpContent bytesContent = new ByteArrayContent(paramFileBytes);
    using (var client = new HttpClient())
    using (var formData = new MultipartFormDataContent())
    {
        formData.Add(stringContent, "param1", "param1");
        formData.Add(fileStreamContent, "file1", "file1");
        formData.Add(bytesContent, "file2", "file2");
        var response = await client.PostAsync(actionUrl, formData);
        if (!response.IsSuccessStatusCode)
        {
            return null;
        }
        return await response.Content.ReadAsStreamAsync();
    }
}

Answer 2:

只发送原始文件：

using(WebClient client = new WebClient()) {
    client.UploadFile(address, filePath);
}

如果要模拟的浏览器的形式与<input type="file"/>那么就是更难。 见这个答案的多/ form-data的答案。

Answer 3:

对我来说client.UploadFile还裹着multipart请求内容，所以我不得不这样做是这样的：

using (WebClient client = new WebClient())
{
    client.Headers.Add("Content-Type", "application/octet-stream");
    using (Stream fileStream = File.OpenRead(filePath))
    using (Stream requestStream = client.OpenWrite(new Uri(fileUploadUrl), "POST"))
    {
        fileStream.CopyTo(requestStream);
    }
}

Answer 4:

我已经得到了同样的问题，这下面的代码完美地回答这个问题：

//Identificate separator
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
//Encoding
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

//Creation and specification of the request
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url); //sVal is id for the webService
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;

string sAuthorization = "login:password";//AUTHENTIFICATION BEGIN
byte[] toEncodeAsBytes = System.Text.ASCIIEncoding.ASCII.GetBytes(sAuthorization);
string returnValue = System.Convert.ToBase64String(toEncodeAsBytes);
wr.Headers.Add("Authorization: Basic " + returnValue); //AUTHENTIFICATION END
Stream rs = wr.GetRequestStream();


string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}"; //For the POST's format

//Writting of the file
rs.Write(boundarybytes, 0, boundarybytes.Length);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(Server.MapPath("questions.pdf"));
rs.Write(formitembytes, 0, formitembytes.Length);

rs.Write(boundarybytes, 0, boundarybytes.Length);

string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
string header = string.Format(headerTemplate, "file", "questions.pdf", contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);

FileStream fileStream = new FileStream(Server.MapPath("questions.pdf"), FileMode.Open, FileAccess.Read);
byte[] buffer = new byte[4096];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
    rs.Write(buffer, 0, bytesRead);
}
fileStream.Close();

byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
rs.Write(trailer, 0, trailer.Length);
rs.Close();
rs = null;

WebResponse wresp = null;
try
{
    //Get the response
    wresp = wr.GetResponse();
    Stream stream2 = wresp.GetResponseStream();
    StreamReader reader2 = new StreamReader(stream2);
    string responseData = reader2.ReadToEnd();
}
catch (Exception ex)
{
    string s = ex.Message;
}
finally
{
    if (wresp != null)
    {
        wresp.Close();
        wresp = null;
    }
    wr = null;
}

Answer 5:

您需要将文件写入请求流：

using (var reqStream = req.GetRequestStream()) 
{    
    reqStream.Write( ... ) // write the bytes of the file
}

Answer 6:

要发布文件，从字节数组：

private static string UploadFilesToRemoteUrl(string url, IList<byte[]> files, NameValueCollection nvc) {

        string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");

        var request = (HttpWebRequest) WebRequest.Create(url);
        request.ContentType = "multipart/form-data; boundary=" + boundary;
        request.Method = "POST";
        request.KeepAlive = true;
        var postQueue = new ByteArrayCustomQueue();

        var formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";

        foreach (string key in nvc.Keys) {
            var formitem = string.Format(formdataTemplate, key, nvc[key]);
            var formitembytes = Encoding.UTF8.GetBytes(formitem);
            postQueue.Write(formitembytes);
        }

        var headerTemplate = "\r\n--" + boundary + "\r\n" +
            "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n" + 
            "Content-Type: application/zip\r\n\r\n";

        var i = 0;
        foreach (var file in files) {
            var header = string.Format(headerTemplate, "file" + i, "file" + i + ".zip");
            var headerbytes = Encoding.UTF8.GetBytes(header);
            postQueue.Write(headerbytes);
            postQueue.Write(file);
            i++;
        }

        postQueue.Write(Encoding.UTF8.GetBytes("\r\n--" + boundary + "--"));

        request.ContentLength = postQueue.Length;

        using (var requestStream = request.GetRequestStream()) {
            postQueue.CopyToStream(requestStream);
            requestStream.Close();
        }

        var webResponse2 = request.GetResponse();

        using (var stream2 = webResponse2.GetResponseStream())
        using (var reader2 = new StreamReader(stream2)) {

            var res =  reader2.ReadToEnd();
            webResponse2.Close();
            return res;
        }
    }

public class ByteArrayCustomQueue {

    private LinkedList<byte[]> arrays = new LinkedList<byte[]>();

    /// <summary>
    /// Writes the specified data.
    /// </summary>
    /// <param name="data">The data.</param>
    public void Write(byte[] data) {
        arrays.AddLast(data);
    }

    /// <summary>
    /// Gets the length.
    /// </summary>
    /// <value>
    /// The length.
    /// </value>
    public int Length { get { return arrays.Sum(x => x.Length); } }

    /// <summary>
    /// Copies to stream.
    /// </summary>
    /// <param name="requestStream">The request stream.</param>
    /// <exception cref="System.NotImplementedException"></exception>
    public void CopyToStream(Stream requestStream) {
        foreach (var array in arrays) {
            requestStream.Write(array, 0, array.Length);
        }
    }
}

Answer 7:

     public string SendFile(string filePath)
            {
                WebResponse response = null;
                try
                {
                    string sWebAddress = "Https://www.address.com";

                    string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
                    byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
                    HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(sWebAddress);
                    wr.ContentType = "multipart/form-data; boundary=" + boundary;
                    wr.Method = "POST";
                    wr.KeepAlive = true;
                    wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
                    Stream stream = wr.GetRequestStream();
                    string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";

                    stream.Write(boundarybytes, 0, boundarybytes.Length);
                    byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(filePath);
                    stream.Write(formitembytes, 0, formitembytes.Length);
                    stream.Write(boundarybytes, 0, boundarybytes.Length);
                    string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
                    string header = string.Format(headerTemplate, "file", Path.GetFileName(filePath), Path.GetExtension(filePath));
                    byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
                    stream.Write(headerbytes, 0, headerbytes.Length);

                    FileStream fileStream = new FileStream(filePath, FileMode.Open, FileAccess.Read);
                    byte[] buffer = new byte[4096];
                    int bytesRead = 0;
                    while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
                        stream.Write(buffer, 0, bytesRead);
                    fileStream.Close();

                    byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
                    stream.Write(trailer, 0, trailer.Length);
                    stream.Close();

                    response = wr.GetResponse();
                    Stream responseStream = response.GetResponseStream();
                    StreamReader streamReader = new StreamReader(responseStream);
                    string responseData = streamReader.ReadToEnd();
                    return responseData;
                }
                catch (Exception ex)
                {
                    return ex.Message;
                }
                finally
                {
                    if (response != null)
                        response.Close();
                }
            }

文章来源: Send a file via HTTP POST with C#