【发布时间】：2010-10-31 15:19:31
【问题描述】：
我目前正在将文件从 C# 应用程序发布到图像主机（KalleLoad.net - 显然是在所有者同意的情况下）。
我已经收到了工作请求的实际发布，但它没有返回我所期望的。上传站点的所有者为我提供了一个 API（某种），如果我将数据发布到某个地址，它将返回一些带有 URL 的 XML。我可以成功发布数据并从服务器获得响应，但是它只是返回主页的代码而不是 XML。我不明白为什么会这样。
我还尝试将数据发布到本地服务器上的简单 PHP 页面，它也返回页面的代码，而不是我指示页面在发布时返回的代码。
以下是我当前发送数据的全部课程。我还一直在比较我从我的应用程序发送的标头与过去半小时 Firefox 发送的标头，我看不出它们之间没有真正改变游戏规则的差异（据我所知）。
在这方面的任何帮助都会非常好，并且会受到欢迎。
问候， 安迪·亨特
using System;
using System.Net;
using System.Text;
using System.IO;
using System.Collections;
using System.Collections.Generic;
using System.Drawing;
namespace Skimpt_3._0
{
class PostFile
{
private Hashtable FormElements;
private HttpWebRequest Request;
private MemoryStream FileStream;
private string CONTENT_BOUNDARY = "---------------------------265001916915724";
public string ContentMIMEType;
public string FormURL;
public string FileName;
public string Response;
public string FileBoxName;
//private int BufferSize;
public PostFile(string Url, string strFileName)
{
FormElements = new Hashtable();
FormURL = Url;
Request = (HttpWebRequest)WebRequest.Create(Url);
//BufferSize = 10240;
FileStream = new MemoryStream();
FileName = strFileName;
}
public void Send(Image image)
{
//Assign the request here too, just in case
Request = (HttpWebRequest)WebRequest.Create(FormURL);
Request.Method = "POST";
Request.AllowWriteStreamBuffering = true;
Request.ProtocolVersion = HttpVersion.Version11;
Request.Headers.Add("Cache-Control", "no-cache");
Request.KeepAlive = true;
Request.ContentType = "multipart/form-data; boundary=---------------------------265001916915724";
StartFileStream(FileStream);
//Must be done in this order for stream to write properly:
//----
//Form elements
//File header
//Image
//File trailer
//----
WriteStringToStream(FileStream, GetFormElements());
WriteImageToStream(FileStream, image, FileName);
CloseStream(FileStream);
byte[] FileByteArray = FileStream.ToArray();
Request.ContentLength = FileByteArray.Length;
Stream PostingStream = Request.GetRequestStream();
PostingStream.Write(FileByteArray, 0, FileByteArray.Length);
WebResponse resp = (HttpWebResponse)Request.GetResponse();
StreamReader SR = new StreamReader(resp.GetResponseStream());
PostingStream.Close();
FileStream.Close();
Request.GetRequestStream().Close();
Response = SR.ReadToEnd();
Request = null;
}
private void CloseStream(MemoryStream FileStream)
{
byte[] BytesToWrite = Encoding.ASCII.GetBytes(CONTENT_BOUNDARY);
FileStream.Write(BytesToWrite, 0, BytesToWrite.Length);
}
private void StartFileStream(MemoryStream FileStream)
{
// \r\n = new line
string str = "POST " + FormURL +"Content-Type: multipart/form-data; boundary="+CONTENT_BOUNDARY+" \r\n \r\n" + CONTENT_BOUNDARY;
byte[] BytesToWrite = Encoding.ASCII.GetBytes(str);
FileStream.Write(BytesToWrite, 0, BytesToWrite.Length);
}
private Byte[] ConvertImageToByteArray(Image img)
{
//Method taken from http://www.csharp-station.com/Articles/Thumbnails.aspx and adapted
MemoryStream memStream = new MemoryStream();
img.Save(memStream, System.Drawing.Imaging.ImageFormat.Png);
byte[] byteArray = new Byte[memStream.Length];
memStream.Position = 0;
memStream.Read(byteArray, 0, (int)memStream.Length);
return byteArray;
}
public void AddFormElement(string ElementName, string ElementValue)
{
FormElements[ElementName] = ElementValue;
}
private string GetFormElements()
{
string str = "";
IDictionaryEnumerator myEnumerator = FormElements.GetEnumerator();
while (myEnumerator.MoveNext())
{
str += CONTENT_BOUNDARY + "\r\n" +
"Content-Disposition: form-data; name=" + myEnumerator.Key +
"\r\n\r\n" +
myEnumerator.Value +"\r\n";
}
return str;
}
private void WriteStringToStream(System.IO.MemoryStream stream, string String)
{
byte[] PostData = System.Text.Encoding.ASCII.GetBytes(String);
stream.Write(PostData, 0, PostData.Length);
}
private void WriteImageToStream(System.IO.MemoryStream Stream, Image img, string FileName)
{
byte[] ByteArray = ConvertImageToByteArray(img);
string head = CONTENT_BOUNDARY + "\r\n" +
"Content-Disposition: form-data; name=\"" + FileBoxName + "\"; filename=\"" + FileName + "\"\r\n" +
"Content-Type: " + ContentMIMEType + "\r\n\r\n";
byte[] header = Encoding.ASCII.GetBytes(head);
Stream.Write(header, 0, header.Length);
Stream.Write(ByteArray, 0, ByteArray.Length);
}
}
}
【问题讨论】：
-
我现在正在阅读您的代码，我确实注意到了一些与网络响应无关的内容。对于您将图像转换为 byte[] 的方法，在执行 img.Save(....) 后，您可以简单地返回 memStream.ToArray()，它将返回图像的 byte[]。
-
谢谢，我现在已经解决了。我不会费心在上面的代码中更新它，因为它真的不需要更新
标签： c# upload screen-scraping