Senden von Dateien mit HTTP-POST in C#

Question

Ich habe eine kleine C# -Web-Anwendung.Wie bekomme ich den C# -Code, der Benutzer ermöglicht, Dateien per HTTP-POST zu senden.Es sollte in der Lage sein, Textdateien, Bilddateien, Excel, CSV zu senden , doc (alle Arten von Dateien) ohne Stream Reader und alle zu verwenden.

Answer 1

können Sie versuchen, den folgenden Code:

public void PostMultipleFiles(string url, string[] files) 
{ 
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x"); 
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(url); 
    httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary; 
    httpWebRequest.Method = "POST"; 
    httpWebRequest.KeepAlive = true; 
    httpWebRequest.Credentials = System.Net.CredentialCache.DefaultCredentials; 
    Stream memStream = new System.IO.MemoryStream(); 
    byte[] boundarybytes =System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary  +"\r\n"); 
    string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}"; 
    string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n"; 
    memStream.Write(boundarybytes, 0, boundarybytes.Length); 
    for (int i = 0; i < files.Length; i++) 
    { 
     string header = string.Format(headerTemplate, "file" + i, files[i]); 
     //string header = string.Format(headerTemplate, "uplTheFile", files[i]); 
     byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header); 
     memStream.Write(headerbytes, 0, headerbytes.Length); 
     FileStream fileStream = new FileStream(files[i], FileMode.Open, 
     FileAccess.Read); 
     byte[] buffer = new byte[1024]; 
     int bytesRead = 0; 
     while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0) 
     { 
      memStream.Write(buffer, 0, bytesRead); 
     } 
     memStream.Write(boundarybytes, 0, boundarybytes.Length); 
     fileStream.Close(); 
    } 
    httpWebRequest.ContentLength = memStream.Length; 
    Stream requestStream = httpWebRequest.GetRequestStream(); 
    memStream.Position = 0; 
    byte[] tempBuffer = new byte[memStream.Length]; 
    memStream.Read(tempBuffer, 0, tempBuffer.Length); 
    memStream.Close(); 
    requestStream.Write(tempBuffer, 0, tempBuffer.Length); 
    requestStream.Close(); 
    try 
    { 
     WebResponse webResponse = httpWebRequest.GetResponse(); 
     Stream stream = webResponse.GetResponseStream(); 
     StreamReader reader = new StreamReader(stream); 
     string var = reader.ReadToEnd(); 

    } 
    catch (Exception ex) 
    { 
     response.InnerHtml = ex.Message; 
    } 
    httpWebRequest = null; 
} 

Answer 2

versuchen diese

string fileToUpload = @"c:\user\test.txt"; 
string url = "http://example.com/upload"; 
using (var client = new WebClient()) 
{ 
byte[] result = client.UploadFile(url, fileToUpload); 
string responseAsString = Encoding.Default.GetString(result); 
} 

Answer 3

.NET 4.5 (von oder .NET 4.0 Hinzufügen des Microsoft.Net.Http Paket von NuGet) Mit Hilfe gibt es einen einfacheren Weg zu Simulieren von Formularanforderungen Hier ein Beispiel:

private 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 = client.PostAsync(actionUrl, formData).Result; 
     if (!response.IsSuccessStatusCode) 
     { 
      return null; 
     } 
     return response.Content.ReadAsStreamAsync().Result; 
    } 
}