C＃：System.Diagnostics.Process.Start（“Explorer.exe”，@“/ select”+ FilePath）。文件名为unicode字符时无法打开文件

Question

I want to open file's location with window Explorer. I am using C# with code

System.Diagnostics.Process.Start("Explorer.exe", @"/select," + FilePath)

it works well with simple English character, but it could not open the file's location if the file's name is Unicode character (Thia language).

Anyone could help please?

Answer 1

试着把它放在引号中：

System.Diagnostics.Process.Start("Explorer.exe", @"/select,""" + FilePath + "\"")

Answer 2

No trouble with this code snippet:

    static void Main(string[] args) {
        string path = @"c:\temp\លួចស្រលាញ់សង្សារគេ.DAT";
        System.IO.File.WriteAllText(path, "hello");
        string txt = System.IO.File.ReadAllText(path);
    }

Windows 7, the file is created and displays correctly in Explorer. You didn't document your operating system version so that's one failure mode, albeit a very small one. Much more likely is trouble with the file system that's mapped to your E: drive. Like a FAT32 volume on a flash drive or a network redirector. Ask questions about that, respectively, at superuser.com and serverfault.com. Do not forget to document those essential details.

Answer 3

Explorer will go to a default folder in this case 'My Documents', if the folder you are attempting to open is not there. Make sure it exists.

Answer 4

Here's the deal as far as I've determined: at least as of Windows 8.1, "Explorer.exe" appears to strip out all combining characters before looking for the file. You can test this either in c# or a console (do chcp 65001 first to get in unicode mode). If you try to open a target named ปู (thai for "crab") it won't work, but if you remove the vowel mark under so that you have just ป, it will work. Further, if you have a folder named ป and you as it to open ปู, it will open the ป folder!

This explains why some other devs had no problem; the problem is not non-ascii : rather, it is filenames with composable characters. Not all languages use them, and even in languages that do, not all file names have them.

The good news is, there's a different way to open these that doesn't have this problem, which is described by @bert-huijben in this answer .

For completeness, here's the version similar to what I ended up using:

    [DllImport("shell32.dll", ExactSpelling = true)]
    public static extern void ILFree(IntPtr pidlList);

    [DllImport("shell32.dll", CharSet = CharSet.Unicode, ExactSpelling = true)]
    public static extern IntPtr ILCreateFromPathW(string pszPath);

    [DllImport("shell32.dll", ExactSpelling = true)]
    public static extern int SHOpenFolderAndSelectItems(IntPtr pidlList, uint cild, IntPtr children, uint dwFlags);

    public void SelectItemInExplorer(string path)
    {
        var pidlList = ILCreateFromPathW(path);
        if(pidlList == IntPtr.Zero)
            throw new Exception(string.Format("ILCreateFromPathW({0}) failed",path));
        try
        {
            Marshal.ThrowExceptionForHR(SHOpenFolderAndSelectItems(pidlList, 0, IntPtr.Zero, 0));
        }
        finally
        {
            ILFree(pidlList);
        }
    }

Answer 5

The following code works for me with files with korean characters (are unicode characters). Please try it and let me know if it works.

       ...
       if (this.IsDirectory())
       {
          OpenFileWith("explorer.exe", this.FullPath, "/root,");
       }
       else
       {
          OpenFileWith("explorer.exe", this.FullPath, "/select,");
       }
      ...

    public static void OpenFileWith(string exePath, string path, string arguments)
    {
        if (path == null)
            return;

        try 
        {
            System.Diagnostics.Process process = new System.Diagnostics.Process();
            process.StartInfo.WorkingDirectory = Path.GetDirectoryName(path);
            if (exePath != null)
            {
                process.StartInfo.FileName = exePath;
                //Pre-post insert quotes for fileNames with spaces.
                process.StartInfo.Arguments = string.Format("{0}\"{1}\"", arguments, path);
            }
            else
            {
                process.StartInfo.FileName = path;
                process.StartInfo.WorkingDirectory = Path.GetDirectoryName(path);
            }
            if (!path.Equals(process.StartInfo.WorkingDirectory))
            {
                process.Start();
            }
        }
        catch(System.ComponentModel.Win32Exception ex) 
        {
            FormManager.DisplayException(ex, MessageBoxIcon.Information);
        }
    }