c# - GetFiles with multiple extensions

Question

Possible Duplicate:
Can you call Directory.GetFiles() with multiple filters?

How do you filter on more than one extension?

I've tried:

FileInfo[] Files = dinfo.GetFiles("*.jpg;*.tiff;*.bmp");
FileInfo[] Files = dinfo.GetFiles("*.jpg,*.tiff,*.bmp");

Answer 1

为什么不创建扩展方法？这更具可读性。

public static IEnumerable<FileInfo> GetFilesByExtensions(this DirectoryInfo dir, params string[] extensions)
{
    if (extensions == null) 
         throw new ArgumentNullException("extensions");
    IEnumerable<FileInfo> files = Enumerable.Empty<FileInfo>();
    foreach(string ext in extensions)
    {
       files = files.Concat(dir.GetFiles(ext));
    }
    return files;
}

编辑：更有效的版本：

public static IEnumerable<FileInfo> GetFilesByExtensions(this DirectoryInfo dir, params string[] extensions)
{
    if (extensions == null) 
         throw new ArgumentNullException("extensions");
    IEnumerable<FileInfo> files = dir.EnumerateFiles();
    return files.Where(f => extensions.Contains(f.Extension));
}

用法：

DirectoryInfo dInfo = new DirectoryInfo(@"c:\MyDir");
dInfo.GetFilesByExtensions(".jpg",".exe",".gif");

Answer 2

您可以获取每个文件，然后过滤数组：

public static IEnumerable<FileInfo> GetFilesByExtensions(this DirectoryInfo dirInfo, params string[] extensions)
{
    var allowedExtensions = new HashSet<string>(extensions, StringComparer.OrdinalIgnoreCase);

    return dirInfo.EnumerateFiles()
                  .Where(f => allowedExtensions.Contains(f.Extension));
}

这将（略微）比这里的所有其他答案更快。
在 .Net 3.5 中，替换EnumerateFiles为GetFiles（速度较慢）。

并像这样使用它：

var files = new DirectoryInfo(...).GetFilesByExtensions(".jpg", ".mov", ".gif", ".mp4");

Answer 3

你不能这样做，因为GetFiles只接受一个搜索模式。相反，您可以GetFiles不使用模式调用，并在代码中过滤结果：

string[] extensions = new[] { ".jpg", ".tiff", ".bmp" };

FileInfo[] files =
    dinfo.GetFiles()
         .Where(f => extensions.Contains(f.Extension.ToLower()))
         .ToArray();

如果您使用的是 .NET 4，则可以使用该EnumerateFiles方法避免一次将所有 FileInfo 对象加载到内存中：

string[] extensions = new[] { ".jpg", ".tiff", ".bmp" };

FileInfo[] files =
    dinfo.EnumerateFiles()
         .Where(f => extensions.Contains(f.Extension.ToLower()))
         .ToArray();

Answer 4

You can use LINQ Union method:

dir.GetFiles("*.txt").Union(dir.GetFiles("*.jpg")).ToArray();

Answer 5

以下检索 jpg、tiff 和 bmp 文件，并为您提供IEnumerable<FileInfo>可以迭代的内容：

var files = dinfo.GetFiles("*.jpg")
    .Concat(dinfo.GetFiles("*.tiff"))
    .Concat(dinfo.GetFiles("*.bmp"));

如果你真的需要一个数组，只需坚持.ToArray()到最后。

Answer 6

我不确定这是否可能。MSDN GetFiles参考说的是搜索模式，而不是搜索模式列表。

我可能倾向于单独获取每个列表并将它们“foreach”到最终列表中。

Answer 7

我知道有一种更优雅的方法可以做到这一点，我愿意接受建议......这就是我所做的：

          try
            {


             // Set directory for list to be made of
                DirectoryInfo jpegInfo = new DirectoryInfo(destinationFolder);
                DirectoryInfo jpgInfo = new DirectoryInfo(destinationFolder);
                DirectoryInfo gifInfo = new DirectoryInfo(destinationFolder);
                DirectoryInfo tiffInfo = new DirectoryInfo(destinationFolder);
                DirectoryInfo bmpInfo = new DirectoryInfo(destinationFolder);

                // Set file type
                FileInfo[] Jpegs = jpegInfo.GetFiles("*.jpeg");
                FileInfo[] Jpgs = jpegInfo.GetFiles("*.jpg");
                FileInfo[] Gifs = gifInfo.GetFiles("*.gif");
                FileInfo[] Tiffs = gifInfo.GetFiles("*.tiff");
                FileInfo[] Bmps = gifInfo.GetFiles("*.bmp");

        //  listBox1.Items.Add(@"");  // Hack for the first list item no preview problem
        // Iterate through each file, displaying only the name inside the listbox...
        foreach (FileInfo file in Jpegs)
        {
                listBox1.Items.Add(file.Name);
                Photo curPhoto = new Photo();
                curPhoto.PhotoLocation = file.FullName;
                metaData.AddPhoto(curPhoto);
            }

          foreach (FileInfo file in Jpgs)
          {
              listBox1.Items.Add(file.Name);
                Photo curPhoto = new Photo();
                curPhoto.PhotoLocation = file.FullName;
                metaData.AddPhoto(curPhoto);
            }
          foreach (FileInfo file in Gifs)
          {
              listBox1.Items.Add(file.Name);
              Photo curPhoto = new Photo();
              curPhoto.PhotoLocation = file.FullName;
              metaData.AddPhoto(curPhoto);
          }
          foreach (FileInfo file in Tiffs)
          {
              listBox1.Items.Add(file.Name);
              Photo curPhoto = new Photo();
              curPhoto.PhotoLocation = file.FullName;
              metaData.AddPhoto(curPhoto);
          }
          foreach (FileInfo file in Bmps)
          {
              listBox1.Items.Add(file.Name);
              Photo curPhoto = new Photo();
              curPhoto.PhotoLocation = file.FullName;
              metaData.AddPhoto(curPhoto);
          }