C# treeView 是否有代码可以将子节点安排为其父节点的最后一个子节点

Question

我正在使用treeView在表单上的treeView中显示序列目录及其子目录和文件，我使用以下方法加载树视图

在表单加载中：

        treeView1.Nodes.Clear();
        toolTip1.ShowAlways = true;
        LoadDirectory("C:\\Windows\\System32\\" + inventedName );  

以及以下3种加载目录和子目录及文件的方法

    public void LoadDirectory(string Dir)
    {

        DirectoryInfo di = new DirectoryInfo(Dir);

        TreeNode tds = treeView1.Nodes.Add(di.Name);

        tds.Tag = di.FullName;
        //tds.StateImageIndex = 0;
        tds.ImageIndex = 0;
        tds.StateImageIndex = 0;
        tds.SelectedImageIndex = 0;
        LoadFiles(Dir, tds);
        LoadSubDirectories(Dir, tds);
    }

    private void LoadSubDirectories(string dir, TreeNode td)
    {

        string[] subdirectoryEntries = Directory.GetDirectories(dir);          

        // Loop through them to see if they have any other subdirectories  
        foreach (string subdirectory in subdirectoryEntries)
        {

            DirectoryInfo di = new DirectoryInfo(subdirectory);
            TreeNode tds = td.Nodes.Add(di.Name);
            renameNodes(tds);    
            //tds.StateImageIndex = 0;
            tds.Tag = di.FullName;
            tds.ImageIndex = 0;
            tds.StateImageIndex = 0;
            tds.SelectedImageIndex = 0;
            LoadFiles(subdirectory, tds);
            LoadSubDirectories(subdirectory, tds);

        }
    }

    private void LoadFiles(string dir, TreeNode td)
    {
        string[] Files = Directory.GetFiles(dir, "*.pdf");

        // Loop through them to see files  
        foreach (string file in Files)
        {
            FileInfo fi = new FileInfo(file);
            TreeNode tds = td.Nodes.Add(fi.Name);
            tds.Tag = fi.FullName;
            tds.ImageIndex = 1;
            tds.StateImageIndex = 1;
            tds.SelectedImageIndex = 1;

        }
    }

我的问题是子目录（文件夹）有特定的名称，我无法更改它 例如：

> root 
    > parent 
          > 1.0 xxx
          > 1.10 xxx
          > 1.2 xxx
          > 1.3 xxx 
          > 1.4 xxx
          > 1.5 xxx
          > 1.6 xxx
          > 1.7 xxx
          > 1.8 xxx
          > 1.9 xxx

但我需要它是那样的

> root 
    > parent 
         > 1.0 xxx
         > 1.2 xxx
         > 1.3 xxx 
         > 1.4 xxx
         > 1.5 xxx
         > 1.6 xxx
         > 1.7 xxx
         > 1.8 xxx
         > 1.9 xxx
         > 1.10 xxx 

愚蠢的（1.10 xxx）孩子必须在（1.9 xxx）孩子之后 正如我所说，我不能重命名错误的文件夹是否有任何方法可以将其发送为最后一个孩子

谢谢你帮助我

Answer 1

几周前我使用 IEquable 做了一个非常相似的解决方案。我在下面的代码中对文件名进行了排序以获得正确的解决方案

   public class Test
    {
        private void LoadFiles(string dir, TreeNode td)
        {
            string[] Files = Directory.GetFiles(dir, "*.pdf");
            Files = Files.Select(x => new MySort(x)).OrderBy(x => x).Select(x => x.filename).ToArray();

            // Loop through them to see files  
            foreach (string file in Files)
            {
                FileInfo fi = new FileInfo(file);
                TreeNode tds = td.Nodes.Add(fi.Name);
                tds.Tag = fi.FullName;
                tds.ImageIndex = 1;
                tds.StateImageIndex = 1;
                tds.SelectedImageIndex = 1;

            }
        }
    }
    public class MySort : IComparable
    {
        private string[] splitvalues { get; set; }
        public string filename { get; set; }

        public MySort(string filename)
        {
            this.filename = filename;
            splitvalues = filename.Split(new char[] { '.', ' ' }, StringSplitOptions.RemoveEmptyEntries).ToArray();

        }
        public int CompareTo(object other)
        {
            MySort otherMySort = (MySort)other;
            int min = Math.Min(this.splitvalues.Length, otherMySort.splitvalues.Length);

            for (int i = 0; i < min; i++)
            {
                string a = this.splitvalues[i];
                string b = otherMySort.splitvalues[i];

                if (a != b)
                {

                    int numberA = 0;
                    int numberB = 0;

                    if (int.TryParse(a, out numberA))
                    {
                        if (int.TryParse(b, out numberB))
                        {
                            int z = numberA.CompareTo(numberB);
                            //a & b are numbers
                            return numberA.CompareTo(numberB);
                        }
                        else
                        {
                            //a number b string
                            return -1;
                        }

                    }
                    else
                    {
                        if (int.TryParse(b, out numberB))
                        {
                            //a string b number
                            return 1;
                        }
                        else
                        {
                            // a string b string
                            return a.CompareTo(b);
                        }
                    }
                }
            }
            return splitvalues.Length.CompareTo(otherMySort.splitvalues.Length);

        }
    }

Answer 2

您可以从下面的测试代码中看到代码正常工作。我做了一个小改动，添加了一个 Sort() 方法，以便更容易调用代码。其他一切都是一样的。 ：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] input = { "1.7.1", "1.7.10", "1.7.2", "1.7.3", "1.7.4", "1.7.5", "1.7.6", "1.7.7", "1.7.8", "1.7.9" };
            string[] output = MySort.Sort(input);
        }
    }
    public class MySort : IComparable
    {
        private string[] splitvalues { get; set; }
        public string filename { get; set; }

        public MySort(string filename)
        {
            this.filename = filename;
            splitvalues = filename.Split(new char[] { '.', ' ' }, StringSplitOptions.RemoveEmptyEntries).ToArray();

        }
        public static string[] Sort(string[] input)
        {
            return input.Select(x => new MySort(x)).OrderBy(x => x).Select(x => x.filename).ToArray();
        }
        public int CompareTo(object other)
        {
            MySort otherMySort = (MySort)other;
            int min = Math.Min(this.splitvalues.Length, otherMySort.splitvalues.Length);

            for (int i = 0; i < min; i++)
            {
                string a = this.splitvalues[i];
                string b = otherMySort.splitvalues[i];

                if (a != b)
                {

                    int numberA = 0;
                    int numberB = 0;

                    if (int.TryParse(a, out numberA))
                    {
                        if (int.TryParse(b, out numberB))
                        {
                            int z = numberA.CompareTo(numberB);
                            //a & b are numbers
                            return numberA.CompareTo(numberB);
                        }
                        else
                        {
                            //a number b string
                            return -1;
                        }

                    }
                    else
                    {
                        if (int.TryParse(b, out numberB))
                        {
                            //a string b number
                            return 1;
                        }
                        else
                        {
                            // a string b string
                            return a.CompareTo(b);
                        }
                    }
                }
            }
            return splitvalues.Length.CompareTo(otherMySort.splitvalues.Length);

        }
    }
}