如何从文件名中提取Guid?

时间:2015-10-27 04:37:36

标签: c# regex guid

我想从下面的字符串数组中提取Guid,并希望输出如下:

Output:Log.txt,Logging.txt

string[] str = new string[] { "1250a2d5-cd40-4bcc-a979-9d6f2cd62b9fLog.txt", "bdb31966-e3c4-4344-b02c-305c0eb0fa0aLogging.txt" }; //file name are Log.txt and Logging.txt
List<string> list= new List<string>();
foreach (var item in str)
{
    if (!string.IsNullOrEmpty(item))
    {
        list.Add();  // here i want to store Log.txt and Logging.txt
    }
}

怎么做?

1 个答案:

答案 0 :(得分:2)

var str = new string[] { "1250a2d5-cd40-4bcc-a979-9d6f2cd62b9fLog.txt", "bdb31966-e3c4-4344-b02c-305c0eb0fa0aLogging.txt" }; //file name are Log.txt and Logging.txt

var arr = str.Select(s=>s.Substring(36)).Where(s=>s.Any()).ToArray();

您可以在此处看到它:https://dotnetfiddle.net/eHy6Fo