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
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。
其他回答
简单的解决方案
string dir = @"D:\PATH";
DateTime from_date = DateTime.Now.Date;
DateTime to_date = DateTime.Now.Date.AddHours(23);
var files = Directory.EnumerateFiles(dir, "*.*",SearchOption.AllDirectories).Select(i=>new FileInfo(i))
.Where(file=>file.LastWriteTime >= from_date && file.LastWriteTime <= to_date);
foreach(var fl in files)
Console.WriteLine(fl.FullName);
在Framework 2.0中,你可以使用(它列出根文件夹的文件,这是最好的最流行的答案):
static void DirSearch(string dir)
{
try
{
foreach (string f in Directory.GetFiles(dir))
Console.WriteLine(f);
foreach (string d in Directory.GetDirectories(dir))
{
Console.WriteLine(d);
DirSearch(d);
}
}
catch (System.Exception ex)
{
Console.WriteLine(ex.Message);
}
}
如果您只需要文件名,并且由于我不喜欢这里的大多数解决方案(特性或可读性方面),那么这个懒惰的解决方案如何?
private void Foo()
{
var files = GetAllFiles("pathToADirectory");
foreach (string file in files)
{
// Use can use Path.GetFileName() or similar to extract just the filename if needed
// You can break early and it won't still browse your whole disk since it's a lazy one
}
}
/// <exception cref="T:System.IO.DirectoryNotFoundException">The specified path is invalid (for example, it is on an unmapped drive).</exception>
/// <exception cref="T:System.UnauthorizedAccessException">The caller does not have the required permission.</exception>
/// <exception cref="T:System.IO.IOException"><paramref name="path" /> is a file name.-or-A network error has occurred.</exception>
/// <exception cref="T:System.IO.PathTooLongException">The specified path, file name, or both exceed the system-defined maximum length. For example, on Windows-based platforms, paths must be less than 248 characters and file names must be less than 260 characters.</exception>
/// <exception cref="T:System.ArgumentNullException"><paramref name="path" /> is null.</exception>
/// <exception cref="T:System.ArgumentException"><paramref name="path" /> is a zero-length string, contains only white space, or contains one or more invalid characters as defined by <see cref="F:System.IO.Path.InvalidPathChars" />.</exception>
[NotNull]
public static IEnumerable<string> GetAllFiles([NotNull] string directory)
{
foreach (string file in Directory.GetFiles(directory))
{
yield return file; // includes the path
}
foreach (string subDir in Directory.GetDirectories(directory))
{
foreach (string subFile in GetAllFiles(subDir))
{
yield return subFile;
}
}
}
一个简单干净的解决方案
/// <summary>
/// Scans a folder and all of its subfolders recursively, and updates the List of files
/// </summary>
/// <param name="sFullPath">Full path of the folder</param>
/// <param name="files">The list, where the output is expected</param>
internal static void EnumerateFiles(string sFullPath, List<FileInfo> fileInfoList)
{
try
{
DirectoryInfo di = new DirectoryInfo(sFullPath);
FileInfo[] files = di.GetFiles();
foreach (FileInfo file in files)
fileInfoList.Add(file);
//Scan recursively
DirectoryInfo[] dirs = di.GetDirectories();
if (dirs == null || dirs.Length < 1)
return;
foreach (DirectoryInfo dir in dirs)
EnumerateFiles(dir.FullName, fileInfoList);
}
catch (Exception ex)
{
Logger.Write("Exception in Helper.EnumerateFiles", ex);
}
}
这就是我们如何将文件作为其子目录的FileInfo,
var dir = new DirectoryInfo(rootPath);
FileInfo[] files = dir.GetFiles("*.*", SearchOption.AllDirectories);
推荐文章
- 实体框架核心:在上一个操作完成之前,在此上下文中开始的第二个操作
- 如何为构造函数定制Visual Studio的私有字段生成快捷方式?
- 为什么Visual Studio 2015/2017/2019测试运行器没有发现我的xUnit v2测试
- 如何使用JSON确保字符串是有效的JSON。网
- AppSettings从.config文件中获取值
- 通过HttpClient向REST API发布一个空体
- 如何检查IEnumerable是否为空或空?
- 自动化invokerrequired代码模式
- 没有ListBox。SelectionMode="None",是否有其他方法禁用列表框中的选择?
- 在c#代码中设置WPF文本框的背景颜色
- 在c#中,什么是单子?
- c#和Java中的泛型有什么不同?和模板在c++ ?
- c#线程安全快速(est)计数器
- 如何将此foreach代码转换为Parallel.ForEach?
- 如何在iis7应用程序池中设置。net Framework 4.5版本