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;
}

下面的函数搜索多个以逗号分隔的模式。你也可以指定一个排除，例如:"!web。Config”将搜索所有文件并排除“web.config”。模式可以混合使用。

private string[] FindFiles(string directory, string filters, SearchOption searchOption)
{
    if (!Directory.Exists(directory)) return new string[] { };

    var include = (from filter in filters.Split(new char[] { ',' }, StringSplitOptions.RemoveEmptyEntries) where !string.IsNullOrEmpty(filter.Trim()) select filter.Trim());
    var exclude = (from filter in include where filter.Contains(@"!") select filter);

    include = include.Except(exclude);

    if (include.Count() == 0) include = new string[] { "*" };

    var rxfilters = from filter in exclude select string.Format("^{0}$", filter.Replace("!", "").Replace(".", @"\.").Replace("*", ".*").Replace("?", "."));
    Regex regex = new Regex(string.Join("|", rxfilters.ToArray()));

    List<Thread> workers = new List<Thread>();
    List<string> files = new List<string>();

    foreach (string filter in include)
    {
        Thread worker = new Thread(
            new ThreadStart(
                delegate
                {
                    string[] allfiles = Directory.GetFiles(directory, filter, searchOption);
                    if (exclude.Count() > 0)
                    {
                        lock (files)
                            files.AddRange(allfiles.Where(p => !regex.Match(p).Success));
                    }
                    else
                    {
                        lock (files)
                            files.AddRange(allfiles);
                    }
                }
            ));

        workers.Add(worker);

        worker.Start();
    }

    foreach (Thread worker in workers)
    {
        worker.Join();
    }

    return files.ToArray();

}

用法:

foreach (string file in FindFiles(@"D:\628.2.11", @"!*.config, !*.js", SearchOption.AllDirectories))
            {
                Console.WriteLine(file);
            }

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。

另一种使用Linq的方式，但是不需要返回所有内容并在内存中过滤。

var files = Directory.GetFiles("C:\\path", "*.mp3", SearchOption.AllDirectories).Union(Directory.GetFiles("C:\\path", "*.jpg", SearchOption.AllDirectories));

它实际上是对GetFiles()的2次调用，但我认为这与问题的精神是一致的，并以一个可枚举的形式返回它们。

/// <summary>
/// Returns the names of files in a specified directories that match the specified patterns using LINQ
/// </summary>
/// <param name="srcDirs">The directories to seach</param>
/// <param name="searchPatterns">the list of search patterns</param>
/// <param name="searchOption"></param>
/// <returns>The list of files that match the specified pattern</returns>
public static string[] GetFilesUsingLINQ(string[] srcDirs,
     string[] searchPatterns,
     SearchOption searchOption = SearchOption.AllDirectories)
{
    var r = from dir in srcDirs
            from searchPattern in searchPatterns
            from f in Directory.GetFiles(dir, searchPattern, searchOption)
            select f;

    return r.ToArray();
}

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();