如何拆分复杂的字符串以获取文件名和base64字符串

Question

嗨我有一个来自XML节点的字符串，我需要拆分字符串并在一个字符串中获取“filename =”值，而在其他字符串中的所有基于block的块都能够解码并创建文件

我正在使用Split（“\ n”）但是为每一行创建了一行，这不是我需要的。任何想法？

<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

      JVBERi0xLjUNJeLjz9MNCjI4MzQgMCBvYmoNPDwvTGluZWFyaXplZCAxL0wgMTAxNTkyOS9PIDI4
      MzcvRSAxMDIyOTMvTiA0Mi9UIDEwMTUxMjkvSCBbIDU4MiA1MDZdPj4NZW5kb2JqDSAgICAgICAg
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      --boundaryfIudow==--
    </Attachment>

Answer 1

你去（src是你的源字符串）：

var split = src.Split('\n').Select(p => p.Trim()).ToList();

var filename = split.First(p => p.StartsWith("filename="));
filename = filename.Substring(10, filename.Length - 11);

var base64 = split[split.Count - 2];

Answer 2

如果您已有XMLNode，请使用它。 将其解析为元素。

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