使用C＃将日志文件解析为XML

Question

I need to separate the information below and only take the XML out, I'm trying to figure out the most efficient way to do this. I'm not sure what approach to start with in regards to removing the first 3 lines, and getting the XML DTD. In my head I was thinking that the best approach to this would be to ignore/remove the 3 lines until XML tags are opened with '<' but I wasn't sure if I should put that in a giant string? Honestly anything would be helpful, I'm stuck figuring this bad boy out, and I'm sure it's not going to be as hard as I'm making it out to be, but I am stuck. Thank you very much!

EDIT: This is a .log file

This is the sample Text Document:

VCS (1.0.11.111): [10/9/2015 12:00:02 AM]
POST https://ex.sample.com/samp/x/sample
Content-Type: application/x-www-form-urlencoded
<?xml version="1.0" encoding="UTF-8"?>
    <command name="sample name_" signature="some stuff" address="sample.com">
    <param name="CurrentVersion">1111</param>
    <param name="MotherboardName">Dell Inc. PowerEdge R420</param>
</command>
HTTP/1.1 200 OK

Answer 1

The easiest way would be to get the first index of < and the last index of > substring your file and let the .Net Xml Parser do its work.

But I am not sure if it is the fastest way.

XML Parsing to class has been answered here

Answer 2

What about using some regular expression? Try this:

        var regex = new Regex(@"<\?xml.*\?>(?<Xml>.*)HTTP/", RegexOptions.Singleline);

        var match = regex.Match(inputString);

        if (match.Success)
        {
            var xmlResult = match.Groups["Xml"].Value;
        }

You will have all xml in the variable xmlResult .