如何递归列出所有文件在一个目录和子目录在c# ?


当前回答

注意。net 4.0中内置了基于迭代器的(而不是基于数组的)文件函数:

foreach (string file in Directory.EnumerateFiles(path, "*.*", SearchOption.AllDirectories))
{
    Console.WriteLine(file);
}

现在我会使用如下的方法;如果你不能访问单个子目录,那么内置的递归方法很容易崩溃…;Queue<string>的使用避免了太多的调用堆栈递归,迭代器块避免了我们拥有一个巨大的数组。

static void Main() {
    foreach (string file in GetFiles(SOME_PATH)) {
        Console.WriteLine(file);
    }
}

static IEnumerable<string> GetFiles(string path) {
    Queue<string> queue = new Queue<string>();
    queue.Enqueue(path);
    while (queue.Count > 0) {
        path = queue.Dequeue();
        try {
            foreach (string subDir in Directory.GetDirectories(path)) {
                queue.Enqueue(subDir);
            }
        }
        catch(Exception ex) {
            Console.Error.WriteLine(ex);
        }
        string[] files = null;
        try {
            files = Directory.GetFiles(path);
        }
        catch (Exception ex) {
            Console.Error.WriteLine(ex);
        }
        if (files != null) {
            for(int i = 0 ; i < files.Length ; i++) {
                yield return files[i];
            }
        }
    }
}

其他回答

IEnumerable<string> GetFilesFromDir(string dir) =>
 Directory.EnumerateFiles(dir).Concat(
 Directory.EnumerateDirectories(dir)
          .SelectMany(subdir => GetFilesFromDir(subdir)));

为了避免unauthoredaccessexception,我使用:

var files = GetFiles(@"C:\", "*.*", SearchOption.AllDirectories);
foreach (var file in files)
{
    Console.WriteLine($"{file}");
}

public static IEnumerable<string> GetFiles(string path, string searchPattern, SearchOption searchOption)
{
    var foldersToProcess = new List<string>()
    {
        path
    };

    while (foldersToProcess.Count > 0)
    {
        string folder = foldersToProcess[0];
        foldersToProcess.RemoveAt(0);

        if (searchOption.HasFlag(SearchOption.AllDirectories))
        {
            //get subfolders
            try
            {
                var subfolders = Directory.GetDirectories(folder);
                foldersToProcess.AddRange(subfolders);
            }
            catch (Exception ex)
            {
                //log if you're interested
            }
        }

        //get files
        var files = new List<string>();
        try
        {
            files = Directory.GetFiles(folder, searchPattern, SearchOption.TopDirectoryOnly).ToList();
        }
        catch (Exception ex)
        {
            //log if you're interested
        }

        foreach (var file in files)
        {
            yield return file;
        }
    }
}

我更喜欢使用DirectoryInfo,因为我可以得到FileInfo的,而不仅仅是字符串。

        string baseFolder = @"C:\temp";
        DirectoryInfo di = new DirectoryInfo(baseFolder);

        string searchPattern = "*.xml";

        ICollection<FileInfo> matchingFileInfos = di.GetFiles(searchPattern, SearchOption.AllDirectories)
            .Select(x => x)
            .ToList();

我这样做的情况下,在未来我需要未来过滤..基于FileInfo的属性。

        string baseFolder = @"C:\temp";
        DirectoryInfo di = new DirectoryInfo(baseFolder);

        string searchPattern = "*.xml";

        ICollection<FileInfo> matchingFileInfos = di.GetFiles(searchPattern, SearchOption.AllDirectories)
            .Where(x => x.LastWriteTimeUtc < DateTimeOffset.Now)
            .Select(x => x)
            .ToList();

如果需要,我也可以求助于字符串。(仍然是未来证明过滤器/where-clause子句的东西。

        string baseFolder = @"C:\temp";
        DirectoryInfo di = new DirectoryInfo(baseFolder);

        string searchPattern = "*.xml";

        ICollection<string> matchingFileNames = di.GetFiles(searchPattern, SearchOption.AllDirectories)
            .Select(x => x.FullName)
            .ToList();

请注意,如果您想通过扩展文件,“.”是一个有效的搜索模式。

一个非常简单的解决方案,返回一个文件列表。

    public static List<string> AllFilesInFolder(string folder)
    {
        var result = new List<string>();

        foreach (string f in Directory.GetFiles(folder))
        {
            result.Add(f);
        }

        foreach (string d in Directory.GetDirectories(folder))
        {
            result.AddRange(AllFilesInFolder(d));
        }

        return result;
    }
var d = new DirectoryInfo(@"C:\logs");
var list = d.GetFiles("*.txt").Select(m => m.Name).ToList();