您可以使用 Guid.TryParse方法 [ ^ ]

  string  filename =   acd5604f-5bbd-4e94-9b40-ea0247cdb4ad; 
 Guid guid; 
 if （Guid.TryParse（filename， out  guid））
 {
 //  删除文件 
} 

模式是：

- 8个十六进制数字

- 破折号

- 4个十六进制数字

- 破折号

- 4个十六进制数字

- 破折号

- 4个十六进制数字

- 破折号

- 12个十六进制数字



正则表达式可以轻松捕获：

正则表达式r = nex正则表达式（ @  ^ [0-9a-zA-Z] {8}  -  [0 -9a-ZA-Z] {4}  -  [0-9A-ZA-Z] {4}  -  [0-9A-ZA-Z] {4}  -  [0-9A-ZA-Z] {12} 

string fileName;
foreach （FileInfo fi in dir.GetFiles（））{
fileName = fi。文件名;
if （r.IsMatch（fileName））{ // 这里文件名是GUID
fi.Delete（）;
}
}

due to some code many files of the form
"acd5604f-5bbd-4e94-9b40-ea0247cdb4ad"
have been created in some of the clients' folders. These files are created using some GUID() function and they have no extension.
I needed to remove them, what is the pattern that helps getting them in the easiest way.

you can use Guid.TryParse Method[^]

string filename = "acd5604f-5bbd-4e94-9b40-ea0247cdb4ad";
Guid guid;
if(Guid.TryParse(filename , out guid))
{
   //delete file 
}

The pattern is:
- 8 hexadecimal digits
- dash
- 4 hexadecimal digits
- dash
- 4 hexadecimal digits
- dash
- 4 hexadecimal digits
- dash
- 12 hexadecimal digits

A regular expression could catch it easily:

Regex r = nex Regex(@"^[0-9a-zA-Z]{8}-[0-9a-zA-Z]{4}-[0-9a-zA-Z]{4}-[0-9a-zA-Z]{4}-[0-9a-zA-Z]{12}

" string fileName; foreach (FileInfo fi in dir.GetFiles()) { fileName = fi.FileName; if (r.IsMatch(fileName)) { // Here the file name is a GUID fi.Delete(); } }