如何用正则表达式获取以下字符串中filename=后的文件名，Content-Type:后的内容？

Content-Disposition: form-data; name="file1"; filename="H:\Documents \haha.txt"
Content-Type: text/plain
如何用正则表达式获取以上字符串中filename=后的文件名，Content-Type:后的内容？

(?is)filename=(['"\s]?)([^'"]+)\1.*?content-type:([^:]+)

取分组2，3 就是你想要的

C# code

        string s = @"Content-Disposition: form-data; name=""file1""; filename=""H:\Documents \haha.txt""
Content-Type: text/plain";
        Match match = Regex.Match(s, @"(?is)filename=""([^""]+)""\s*Content-Type:([^;]+)");
        Response.Write(match.Groups[1].Value + "<br/>");
        Response.Write(match.Groups[2].Value + "<br/>");

输出：
H:\Documents \haha.txt
text/plain

private void mnuFileOpen_Click(object sender ,System.EventArgs e)
  {
  // Intialize with the last file name used.
  openFileDialog1.FileName=m_strFileName;
  //Set the filter for text files
  openFileDialog1.Filter=
  "Text files(*.txt)|*.txt|C# files (*.cs)|*.cs";
  //Show the Read Only check box on the dialong box
  openFileDialog1.ShowReadOnly=true;
  //the default extension is for text files
  openFileDialog1.DefaultExt=".txt";
  if (openFileDialog1.ShowDialog()==
  DialogResult.Cancel)