C# 在 post 请求中发送文件+参数
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Send file+parameters in post request
提问by m.qayyum
I'm using this code to send parameters to a webpage and getting correct response from it.
我正在使用此代码将参数发送到网页并从中获得正确的响应。
System.Net.WebClient oWeb = new System.Net.WebClient();
oWeb.Proxy = System.Net.WebRequest.DefaultWebProxy;
oWeb.Proxy.Credentials = System.Net.CredentialCache.DefaultCredentials;
oWeb.Headers.Add("Content-Type", "application/x-www-form-urlencoded");
byte[] bytArguments = System.Text.Encoding.ASCII.GetBytes("value1=123&value2=xyz");
byte[] bytRetData = oWeb.UploadData("http://website.com/file.php", "POST", bytArguments);
response = System.Text.Encoding.ASCII.GetString(bytRetData);
But now I want to send a file like (.doc) to it + above parameters(value1, value2), but I don't know how to do that.
但是现在我想向它发送一个类似 ( .doc)的文件+ 上面的参数 ( value1, value2)，但我不知道该怎么做。
采纳答案by Hamlet Hakobyan
Use WebClient.QueryStringto pass name/value pairs associated with the request.
使用WebClient.QueryString传递与请求关联的名称/值对。
NameValueCollection parameters = new NameValueCollection();
parameters.Add("value1", "123");
parameters.Add("value2", "xyz");
oWeb.QueryString = parameters;
var responseBytes = oWeb.UploadFile("http://website.com/file.php", "path to file");
string response = Encoding.ASCII.GetString(responseBytes);
回答by m.qayyum
public static void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
{
string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");
HttpWebRequest wr = (HttpWebRequest)WebRequest.Create(url);
wr.ContentType = "multipart/form-data; boundary=" + boundary;
wr.Method = "POST";
wr.KeepAlive = true;
wr.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream rs = wr.GetRequestStream();
string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
foreach (string key in nvc.Keys)
{
rs.Write(boundarybytes, 0, boundarybytes.Length);
string formitem = string.Format(formdataTemplate, key, nvc[key]);
byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
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, paramName, file, contentType);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
rs.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(file, 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();
WebResponse wresp = null;
try
{
wresp = wr.GetResponse();
Stream stream2 = wresp.GetResponseStream();
StreamReader reader2 = new StreamReader(stream2);
result = reader2.ReadToEnd();
}
catch (Exception ex)
{
System.Windows.MessageBox.Show("Error occurred while converting file", "Error!");
if (wresp != null)
{
wresp.Close();
wresp = null;
}
}
finally
{
wr = null;
}
}
Copied from SO but can't remember it's link. And this is how it will be used
从 SO 复制，但不记得它的链接。这就是它将如何使用
NameValueCollection nvc = new NameValueCollection();
nvc.Add("parm1", "value1");
nvc.Add("parm2", "value2");
nvc.Add("parm3", "value3");
HttpUploadFile("http://www.example.com/upload.php",@filepath, "file", "text/html", nvc);
Here @filepathis the path of your file you want to upload: c:\file_to_upload.doc
And fileis the name of the filed used in php as $_Files['file']
这@filepath是您要上传的文件的路径：c:\file_to_upload.doc 并且file是在 php 中使用的文件名作为 $_Files['file']
