使用 C # 通过 HTTP POST 发送文件

我一直在到处寻找和阅读，但没有找到任何真正有用的东西。

我正在编写一个小的 C # win 应用程序，它允许用户发送文件到 Web 服务器，不是通过 FTP，而是通过 HTTP 使用 POST。可以把它想象成一个运行在 Windows 应用程序上的 web 表单。

我有我的 HttpWebRequest 对象创建使用这样的东西

HttpWebRequest req = WebRequest.Create(uri) as HttpWebRequest

也设置了 Method，ContentType和 ContentLength的属性。但是我只能做到这些。

这是我的密码:

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)
{
}

因此，我的问题基本上就是如何通过 HTTP POST 使用 C # 发送一个文件(文本文件、图像、音频等)。

谢谢！

305322 次浏览
小开

To send the raw file only:

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

If you want to emulate a browser form with an <input type="file"/>, then that is harder. See this answer for a multipart/form-data answer.

小开

You need to write your file to the request stream:

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

I had got the same problem and this following code answered perfectly at this problem :

//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;
}
小开
最佳答案

Using .NET 4.5 (or .NET 4.0 by adding the Microsoft.Net.Http package from NuGet) there is an easier way to simulate form requests. Here is an example:

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();
}
}
小开

To post files as from byte arrays:

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);
}
}
}
小开

For me client.UploadFile still wrapped the content in a multipart request so I had to do it like this:

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);
}
}
小开 
     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();
}
}
小开

Using .NET 4.5 trying to perform form POST file upload. Tried most of the methods above but to no avail. Found the solution here https://www.c-sharpcorner.com/article/upload-any-file-using-http-post-multipart-form-data

But I am not not keen as I do not understand why we still need to deal with such low level programming in these common usages (should be handled nicely by framework)

小开

You can do it directly with HttpWebRequest/HttpWebResponse like this.

        string serviceUrl = string.Format("{0}/upload?param={1}", "http://127.0.0.1:8080", HttpUtility.UrlEncode(parameter));
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(serviceUrl);
request.Method = "POST";
request.KeepAlive = true;
        

FileStream file = File.OpenRead(pathToFile);
request.ContentLength = file.Length;


file.Seek(0, SeekOrigin.Begin);
file.CopyTo(request.GetRequestStream());


HttpWebResponse response = (request.GetResponse() as HttpWebResponse);
StreamReader reader = new StreamReader(response.GetResponseStream(), Encoding.UTF8);
string responseText = reader.ReadToEnd();

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