还有一个下降解决方案，似乎没有任何内存或性能开销，而且相当优雅:

string[] filters = new[]{"*.jpg", "*.png", "*.gif"};
string[] filePaths = filters.SelectMany(f => Directory.GetFiles(basePath, f)).ToArray();

Using GetFiles search pattern for filtering the extension is not safe!! For instance you have two file Test1.xls and Test2.xlsx and you want to filter out xls file using search pattern *.xls, but GetFiles return both Test1.xls and Test2.xlsx I was not aware of this and got error in production environment when some temporary files suddenly was handled as right files. Search pattern was *.txt and temp files was named *.txt20181028_100753898 So search pattern can not be trusted, you have to add extra check on filenames as well.

您可以将此添加到项目中

public static class Collectables {
    public static List<System.IO.FileInfo> FilesViaPattern(this System.IO.DirectoryInfo fldr, string pattern) {
        var filter = pattern.Split(" ");
        return fldr.GetFiles( "*.*", System.IO.SearchOption.AllDirectories)
            .Where(l => filter.Any(k => l.Name.EndsWith(k))).ToList();
    }
}

然后像这样在任何地方使用它

new System.IO.DirectoryInfo("c:\\test").FilesViaPattern("txt doc any.extension");

只是找到了另一种方法。仍然不是一次操作，而是把它扔出去，看看其他人是怎么想的。

private void getFiles(string path)
{
    foreach (string s in Array.FindAll(Directory.GetFiles(path, "*", SearchOption.AllDirectories), predicate_FileMatch))
    {
        Debug.Print(s);
    }
}

private bool predicate_FileMatch(string fileName)
{
    if (fileName.EndsWith(".mp3"))
        return true;
    if (fileName.EndsWith(".jpg"))
        return true;
    return false;
}

Let

var set = new HashSet<string>(
    new[] { ".mp3", ".jpg" },
    StringComparer.OrdinalIgnoreCase); // ignore case
var dir = new DirectoryInfo(path);

Then

dir.EnumerateFiles("*.*", SearchOption.AllDirectories)
   .Where(f => set.Contains(f.Extension));

or

from file in dir.EnumerateFiles("*.*", SearchOption.AllDirectories)
from ext in set // makes sense only if it's just IEnumerable<string> or similar
where String.Equals(ext, file.Extension, StringComparison.OrdinalIgnoreCase)
select file;

或者你可以直接将扩展名字符串转换为string ^

vector <string>  extensions = { "*.mp4", "*.avi", "*.flv" };
for (int i = 0; i < extensions.size(); ++i)
{
     String^ ext = gcnew String(extensions[i].c_str());;
     String^ path = "C:\\Users\\Eric\\Videos";
     array<String^>^files = Directory::GetFiles(path,ext);
     Console::WriteLine(ext);
     cout << " " << (files->Length) << endl;
}