Tôi đã làm điều này năm ngoái dựa trên một số mã mà tôi đã tìm thấy. Nó hỗ trợ chính xác những gì bạn muốn, cả hai tập tin và giá trị.
Đây là lớp được gọi là HttpForm
:
public class HttpForm {
private Dictionary<string, string> _files = new Dictionary<string, string>();
private Dictionary<string, string> _values = new Dictionary<string, string>();
public HttpForm(string url) {
this.Url = url;
this.Method = "POST";
}
public string Method { get; set; }
public string Url { get; set; }
//return self so that we can chain
public HttpForm AttachFile(string field, string fileName) {
_files[field] = fileName;
return this;
}
public HttpForm ResetForm(){
_files.Clear();
_values.Clear();
return this;
}
//return self so that we can chain
public HttpForm SetValue(string field, string value) {
_values[field] = value;
return this;
}
public HttpWebResponse Submit() {
return this.UploadFiles(_files, _values);
}
private HttpWebResponse UploadFiles(Dictionary<string, string> files, Dictionary<string, string> otherValues) {
var req = (HttpWebRequest)WebRequest.Create(this.Url);
req.Timeout = 10000 * 1000;
req.Accept = "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8";
req.AllowAutoRedirect = false;
var mimeParts = new List<MimePart>();
try {
if (otherValues != null) {
foreach (var fieldName in otherValues.Keys) {
var part = new MimePart();
part.Headers["Content-Disposition"] = "form-data; name=\"" + fieldName + "\"";
part.Data = new MemoryStream(Encoding.UTF8.GetBytes(otherValues[fieldName]));
mimeParts.Add(part);
}
}
if (files != null) {
foreach (var fieldName in files.Keys) {
var part = new MimePart();
part.Headers["Content-Disposition"] = "form-data; name=\"" + fieldName + "\"; filename=\"" + files[fieldName] + "\"";
part.Headers["Content-Type"] = "application/octet-stream";
part.Data = File.OpenRead(files[fieldName]);
mimeParts.Add(part);
}
}
string boundary = "----------" + DateTime.Now.Ticks.ToString("x");
req.ContentType = "multipart/form-data; boundary=" + boundary;
req.Method = this.Method;
long contentLength = 0;
byte[] _footer = Encoding.UTF8.GetBytes("--" + boundary + "--\r\n");
foreach (MimePart part in mimeParts) {
contentLength += part.GenerateHeaderFooterData(boundary);
}
req.ContentLength = contentLength + _footer.Length;
byte[] buffer = new byte[8192];
byte[] afterFile = Encoding.UTF8.GetBytes("\r\n");
int read;
using (Stream s = req.GetRequestStream()) {
foreach (MimePart part in mimeParts) {
s.Write(part.Header, 0, part.Header.Length);
while ((read = part.Data.Read(buffer, 0, buffer.Length)) > 0)
s.Write(buffer, 0, read);
part.Data.Dispose();
s.Write(afterFile, 0, afterFile.Length);
}
s.Write(_footer, 0, _footer.Length);
}
var res = (HttpWebResponse)req.GetResponse();
return res;
} catch (Exception ex) {
Console.WriteLine(ex.Message);
foreach (MimePart part in mimeParts)
if (part.Data != null)
part.Data.Dispose();
return (HttpWebResponse)req.GetResponse();
}
}
private class MimePart {
private NameValueCollection _headers = new NameValueCollection();
public NameValueCollection Headers { get { return _headers; } }
public byte[] Header { get; protected set; }
public long GenerateHeaderFooterData(string boundary) {
StringBuilder sb = new StringBuilder();
sb.Append("--");
sb.Append(boundary);
sb.AppendLine();
foreach (string key in _headers.AllKeys) {
sb.Append(key);
sb.Append(": ");
sb.AppendLine(_headers[key]);
}
sb.AppendLine();
Header = Encoding.UTF8.GetBytes(sb.ToString());
return Header.Length + Data.Length + 2;
}
public Stream Data { get; set; }
}
}
Bạn có thể sử dụng nó như thế này:
var file1 = @"C:\file";
var file2 = @"C:\file2";
var yourUrl = "http://yourdomain.com/process.php";
var httpForm = new HttpForm(yourUrl);
httpForm.AttachFile("file1", file1).AttachFile("file2", file2);
httpForm.setValue("foo", "some foo").setValue("blah", "rarrr!");
httpForm.Submit();
Hãy cho tôi biết nếu nó làm việc cho bạn.
Nguồn
2012-02-04 19:08:05
bản sao có thể có của [Tải lên tệp với HTTPWebyêu cầu (multipart/form-data)] (http://stackoverflow.com/questions/566462/upload-files-with-httpwebrequest-multipart-form-data) –