如何使用正则表达式取出以下字符串中的Base64编码？

[dy630]

+OK 3170 octets
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Date: Mon, 4 Nov 2002 17:8:49 +0800
From: "welcome" <[e-mail]>
To: [e-mail] <[e-mail]>
Subject: fdafdasfda
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This is a multi-part message in MIME format.

--=====000_Dragon136765354575_=====
Content-Type: text/plain;
charset="GB2312"
Content-Transfer-Encoding: quoted-printable

welcome=A3=AC=C4=FA=BA=C3=A3=A1

=09

=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=D6=C2
=C0=F1=A3=A1
=09=09=09=09

=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1welcome
=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=[e-mail]
=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A1=A12002-11-04

--=====000_Dragon136765354575_=====
Content-Type: application/octet-stream;
name="attach.txt"
Content-Transfer-Encoding: base64
Content-Disposition: attachment;
filename="attach.txt"

x+m7sLTzsca5qSANCiAgIA0KICDFrqO6xOPV5rXEsbPK7MHLztK1xNK7x9DC8KO/ztK1xMntuN+h
oszl1tihotfuz7K7trXEus3X7szW0eG1xKOsxOO1ucu1y7W/tKOhIA0KICDE0KO6ye243y4uLi4u
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--=====000_Dragon136765354575_=====--

[dy630]

太牛了！！ 三剑客都来了，感动ing

再问一下，如果有不只一段的base64编码怎么办？(不管这个问题有没回答，下午结贴。thanks)

[spring_ok]

不一定要用正则表达式。如果非要用，对不起，我还没想到好的方法。

先找到：
boundary="=====000_Dragon136765354575_====="
取得分界的字符串。
然后用它 split 整个串。在得到的数组中循环，检查是否包含：base64这个词来确定是否是你要的。

[saucer]

using System;
using System.Text.RegularExpressions;


Regex re = new Regex(@"filename=""(?<filename>[^""]+)""\s+(?<base64>.*)\s+-+=+", RegexOptions.Singleline | RegexOptions.IgnoreCase);
Match m = re.Match(str);
if (m.Success)
{
Console.WriteLine(m.Groups["filename"].Value);
Console.WriteLine(m.Groups["base64"].Value);
}

[qiushuiwuhen]

Content-Transfer-Encoding不一定是base64
http://www.google.com/search?q=Content-Transfer-Encoding+filename+-base64

try

Regex re = new Regex(@"Content-Transfer-Encoding: base64(?<filename>.+?)(\r\n){2}(?<base64>.*?)(\r\n){2}", RegexOptions.Singleline);