C# 如何拆分复杂字符串以获取文件名和base64字符串

您好，我有一个来自XML节点的字符串，我需要拆分该字符串并在一个字符串中获取“filename=”值，在另一个字符串中获取based64的所有块，以便能够对其进行解码并创建文件

我正在使用拆分（“\n”），但正在为每一行创建一行，这不是我需要的。有什么想法吗

<Attachment>
      --boundaryfIudow==
      Content-Type: application/octet-stream;   name="IERL4-12-61.pdf"
      Content-Disposition: attachment; name="IERL4-12-61.pdf";
      filename="IERL4-12-61.pdf"
      Content-Transfer-Encoding: base64

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      --boundaryfIudow==--
    </Attachment>

--边界流==
内容类型：应用程序/八位字节流；name=“IERL4-12-61.pdf”
内容处置：附件；name=“IERL4-12-61.pdf”；
filename=“IERL4-12-61.pdf”
内容传输编码：base64
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--边界流==--

给你（src是你的源字符串）：

---

如果您已经拥有XMLNode，请使用它。 将其解析为一个元素

XmlElement attachment = (XmlElement)node; 
var filename = attachment.GetAttribute("filename").FirstOrDefault();

---

非常感谢，但我在这一行var filename=split.First（p=>p.StartsWith（“filename=”））中收到一个错误“Message=Sequence不包含匹配元素”；但是“filename=”exist..则无法正确分割。请参见“调试”中的“拆分字符串得到的值”。这是我的代码：我在特定的行中得到了这个“filename=\”IERL4-12-61.pdf\”，所以这里有一些空白字符。没什么大不了的，你只需要在做任何事情之前修剪分割的条目。我更新了我的答案。请注意，

.Length

已更改为

.Count

。谢谢，我现在正在学习。。。我没有弄错名字。。但我现在得到的是所有base64块代码，只有第一行..它不是属性，而是元素中的纯文本。

XmlElement attachment = (XmlElement)node; 
var filename = attachment.GetAttribute("filename").FirstOrDefault();