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？

------解决方案--------------------
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)