Nop……我相信你必须拨打尽可能多的你想要的文件类型。

我会自己创建一个函数，在我需要的扩展字符串上使用数组，然后迭代该数组，进行所有必要的调用。该函数将返回与我发送的扩展名匹配的文件的通用列表。

希望能有所帮助。

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;

如果你使用VB。NET(或将依赖导入到你的c#项目中)，实际上存在一个方便的方法，允许过滤多个扩展:

Microsoft.VisualBasic.FileIO.FileSystem.GetFiles("C:\\path", Microsoft.VisualBasic.FileIO.SearchOption.SearchAllSubDirectories, new string[] {"*.mp3", "*.jpg"});

在VB。NET可以通过My-namespace访问:

My.Computer.FileSystem.GetFiles("C:\path", FileIO.SearchOption.SearchAllSubDirectories, {"*.mp3", "*.jpg"})

不幸的是，这些方便的方法不支持像Directory.EnumerateFiles()那样的惰性求值变体。

是什么

string[] filesPNG = Directory.GetFiles(path, "*.png");
string[] filesJPG = Directory.GetFiles(path, "*.jpg");
string[] filesJPEG = Directory.GetFiles(path, "*.jpeg");

int totalArraySizeAll = filesPNG.Length + filesJPG.Length + filesJPEG.Length;
List<string> filesAll = new List<string>(totalArraySizeAll);
filesAll.AddRange(filesPNG);
filesAll.AddRange(filesJPG);
filesAll.AddRange(filesJPEG);

DirectoryInfo directory = new DirectoryInfo(Server.MapPath("~/Contents/"));

//Using Union

FileInfo[] files = directory.GetFiles("*.xlsx")
                            .Union(directory
                            .GetFiles("*.csv"))
                            .ToArray();

List<string> FileList = new List<string>();
DirectoryInfo di = new DirectoryInfo("C:\\DirName");

IEnumerable<FileInfo> fileList = di.GetFiles("*.*");

//Create the query
IEnumerable<FileInfo> fileQuery = from file in fileList
                                  where (file.Extension.ToLower() == ".jpg" || file.Extension.ToLower() == ".png")
                                  orderby file.LastWriteTime
                                  select file;

foreach (System.IO.FileInfo fi in fileQuery)
{
    fi.Attributes = FileAttributes.Normal;
    FileList.Add(fi.FullName);
}