Question

我正在尝试比较2个文件夹结构，并找到不在另一个树中的最高文件夹。

但是当给出一个例子时，这可能会更清楚：说我们有这两个文件夹：

Fo1: sub 0 (fo: 3)
 > Fo: sub 0 sub 0 (fo: 2)
 >  > Fo: sub 0 sub 0 sub 0 (fi: 3)
 >  > Fo: sub 0 sub 0 sub 1 (fi: 3)
 > Fo: sub 0 sub 1 (fo: 1)
 >  > Fo: sub 0 sub 0 sub 0 (fi: 1)
 > Fo: sub 0 sub 2 (fo: 2)
 >  > Fo: sub 0 sub 1 sub 0 (fi: 2)
 >  > Fo: sub 0 sub 1 sub 1 (fi: 1)

Fo2: sub 0 (fo: 3)
 > Fo: sub 0 sub 0 (fo: 1)
 >  > Fo: sub 0 sub 0 sub 0 (fi: 3)
 > Fo: sub 0 sub 1 (fo: 3)
 >  > Fo: sub 0 sub 1 sub 0 (fi: 1)
 >  > Fo: sub 0 sub 1 sub 1
 >  > Fo: sub 0 sub 1 sub 2 (fi: 4)
 > Fo: sub 0 sub 2 (fo: 2)
 >  > Fo: sub 0 sub 2 sub 0 (fi: 1)
 >  > Fo: sub 0 sub 2 sub 1

结果将是：

 Fo: sub 0 sub 1 sub 1
 Fo: sub 0 sub 1 sub 2
 Fo: sub 0 sub 2

因此，例如，由于已添加了根，因此未添加sub 0 sub 2 sub 0和sub 0 sub 2 sub 1。并且sub 0 sub 0树中没有添加任何内容，因为所有内容都已存在

我尝试了这两种方法：

private static bool compareFolders(Folder folder1, Folder folder2, List<Folder> newFolders, int index = 0) {
    // if has a folder that is same or is in root
    if (folder1.title == folder2.title) {
        Console.WriteLine("Same: " + folder1.title);

        // check for childs
        index++;
        if (folder1.childs != null) {
            foreach (Folder sub in folder1.childs) {
                if (folder2.childs != null) {
                    foreach (Folder sub2 in folder2.childs) {
                        compareFolders(sub, sub2, newFolders, index);
                    }
                }
            }
        }
        return true;
    }
    // moment there false, check if root doesn't 
    if (folder2.childs != null && folder2.childs.Count > 0)
        // folder 2 has folder that folder 1 doesn't have
        newFolders.AddRange(folder2.childs);
    return false;
}

这是最好的工作，但我看到每个文件夹都与更多的文件夹进行比较，然后它应该。

所以我尝试切换到lambda，但无法弄清楚如何做到这一点。这就是我得到的：

private static void compareFolders2(Folder folder1, Folder folder2, List<Folder> newFolders, int index = 0) {
    if (folder2.childs != null && folder2.childs.Count > 0) {
        if (folder1.childs != null && folder1.childs.Count > 0) {
            var similarFolders = folder1.childs.FindAll(f => folder2.childs.Select(f2 => f2.title).ToList().Contains(f.title));
            if (similarFolders.Count > 0) {
                foreach (Folder folder in similarFolders) {
                    compareFolders2(folder, folder2.childs.Find(f => f.title == folder.title), newFolders);
                }
            } else {
                newFolders.AddRange(folder2.childs);
            }
        }
    }
}

如果你想尝试一些东西我做了一个简单的C＃控制台应用程序，它有一些测试文件夹结构的生成方法，并像示例一样打印 https://gist.github.com/cskiwi/d2e5002af98da6982971

注意：第二个文件夹并不总是包含需要添加的文件夹，在 otherFolder 上增加sublevels / maxfolders以使其更有可能发生

Answer 1

我想通过这样做来解决它：

    private static void compareFolders(Folder folder1, Folder folder2, List<Folder> newFolders) {
        if (folder2.childs != null) {
            if (folder1.childs != null) {
                // check for folder 2 childs not in folder 1
                newFolders.AddRange(folder2.childs.FindAll(f => !folder1.childs.Select(f2 => f2.title).ToList().Contains(f.title)));
                // for similar folders go one deeper
                folder1.childs.FindAll(f => folder2.childs.Select(f2 => f2.title).ToList().Contains(f.title)).ForEach(f1 => compareFolders3(f1, folder2.childs.Find(f2 => f2.title == f1.title), newFolders));
            } else {
                // folder 1 doesn't have childs so add all of folder2
                newFolders.AddRange(folder2.childs);
            }
        }
    }

仍在检查，但看起来喜欢它正在运作

用于发现差异的lambda表达式

1 个答案: