WebClient上传带有POST值的UploadFile
我想使用WebClient类将文件上传到主机。我还想传递一些应该在服务器部分(PHP)的$ _POST数组中显示的值。我想一口气做
我用过下面的代码
using (WebClient wc = new WebClient()){
wc.Encoding = Encoding.UTF8;
NameValueCollection values = new NameValueCollection();
values.Add("client", "VIP");
values.Add("name", "John Doe");
wc.QueryString = values; // this displayes in $_GET
byte[] ans= wc.UploadFile(address, dumpPath);
}
如果我使用QueryString属性,则$ _GET数组中显示的值。但我想通过post方法发送它
回答:
没有内置功能可让您执行此操作。我已经在博客中发布了可以使用的扩展程序。以下是相关的类:
public class UploadFile{
public UploadFile()
{
ContentType = "application/octet-stream";
}
public string Name { get; set; }
public string Filename { get; set; }
public string ContentType { get; set; }
public Stream Stream { get; set; }
}
public byte[] UploadFiles(string address, IEnumerable<UploadFile> files, NameValueCollection values)
{
var request = WebRequest.Create(address);
request.Method = "POST";
var boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x", NumberFormatInfo.InvariantInfo);
request.ContentType = "multipart/form-data; boundary=" + boundary;
boundary = "--" + boundary;
using (var requestStream = request.GetRequestStream())
{
// Write the values
foreach (string name in values.Keys)
{
var buffer = Encoding.ASCII.GetBytes(boundary + Environment.NewLine);
requestStream.Write(buffer, 0, buffer.Length);
buffer = Encoding.ASCII.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"{1}{1}", name, Environment.NewLine));
requestStream.Write(buffer, 0, buffer.Length);
buffer = Encoding.UTF8.GetBytes(values[name] + Environment.NewLine);
requestStream.Write(buffer, 0, buffer.Length);
}
// Write the files
foreach (var file in files)
{
var buffer = Encoding.ASCII.GetBytes(boundary + Environment.NewLine);
requestStream.Write(buffer, 0, buffer.Length);
buffer = Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"{2}", file.Name, file.Filename, Environment.NewLine));
requestStream.Write(buffer, 0, buffer.Length);
buffer = Encoding.ASCII.GetBytes(string.Format("Content-Type: {0}{1}{1}", file.ContentType, Environment.NewLine));
requestStream.Write(buffer, 0, buffer.Length);
file.Stream.CopyTo(requestStream);
buffer = Encoding.ASCII.GetBytes(Environment.NewLine);
requestStream.Write(buffer, 0, buffer.Length);
}
var boundaryBuffer = Encoding.ASCII.GetBytes(boundary + "--");
requestStream.Write(boundaryBuffer, 0, boundaryBuffer.Length);
}
using (var response = request.GetResponse())
using (var responseStream = response.GetResponseStream())
using (var stream = new MemoryStream())
{
responseStream.CopyTo(stream);
return stream.ToArray();
}
}
现在您可以在应用程序中使用它:
using (var stream = File.Open(dumpPath, FileMode.Open)){
var files = new[]
{
new UploadFile
{
Name = "file",
Filename = Path.GetFileName(dumpPath),
ContentType = "text/plain",
Stream = stream
}
};
var values = new NameValueCollection
{
{ "client", "VIP" },
{ "name", "John Doe" },
};
byte[] result = UploadFiles(address, files, values);
}
现在,在你的PHP脚本,你可以使用$_POST["client"]
,$_POST["name"]
和$_FILES["file"]
。
以上是 WebClient上传带有POST值的UploadFile 的全部内容, 来源链接: utcz.com/qa/426319.html