得到目录大小的更有效的方法 [英] More efficient method of getting Directory size

问题描述

我已经建立一个递归函数来得到一个文件夹路径的目录大小。然而，它的工作原理，与不断增长数量的目录我要通过搜索（以及在每个相应的文件夹的文件的数目），这是一个非常缓慢的，低效率的方法。

I've already build a recursive function to get the directory size of a folder path. It works, however with the growing number of directories I have to search through (and number of files in each respective folder), this is a very slow, inefficient method.

static string GetDirectorySize(string parentDir)
{
    long totalFileSize = 0;

    string[] dirFiles = Directory.GetFiles(parentDir, "*.*", 
                            System.IO.SearchOption.AllDirectories);

    foreach (string fileName in dirFiles)
    {
        // Use FileInfo to get length of each file.
        FileInfo info = new FileInfo(fileName);
        totalFileSize = totalFileSize + info.Length;
    }
    return String.Format(new FileSizeFormatProvider(), "{0:fs}", totalFileSize);
}

这是搜索所有子目录的说法路径，因此 dirFiles 阵列变得相当大。有没有更好的方法来做到这一点？我四处搜查，但没有发现任何东西。

This is searches all subdirectories for the argument path, so the dirFiles array gets quite large. Is there a better method to accomplish this? I've searched around but haven't found anything yet.

另一个想法掠过我脑海里把结果在缓存中，当函数再次被调用，尝试并发现已更改的分歧，只能重新搜索文件夹。不知道这是一件好事，要么...

Another idea that crossed my mind was putting the results in a cache and when the function is called again, try and find the differences and only re-search folders that have changed. Not sure if that's a good thing either...

推荐答案

您先扫描树来获取所有文件的列表。那么你重新打开每个文件来获得它的大小。这相当于扫描两次。

You are first scanning the tree to get a list of all files. Then you are reopening every file to get its size. This amounts to scanning twice.

我建议你使用DirectoryInfo.GetFiles将交给你了FileInfo对象直接。这些对象预填充其长度。

I suggest you use DirectoryInfo.GetFiles which will hand you FileInfo objects directly. These objects are pre-filled with their length.

在.NET 4中，您还可以使用EnumerateFiles方法，该方法将返回一个懒惰IEnumable。

In .NET 4 you can also use the EnumerateFiles method which will return you a lazy IEnumable.

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