Power-Mosfet Power-Mosfet - 3 months ago 11
C# Question

How do I display a file's Properties dialog from C#?

how to open an file's Properties dialog by a button

private void button_Click(object sender, EventArgs e)
{
string path = @"C:\Users\test\Documents\tes.text";
// how to open this propertie
}


Thank you.

For example if want the System properties

Process.Start("sysdm.cpl");


But how do i get the Properties dialog for a file path?

Answer

Solution is:

using System.Runtime.InteropServices;

[DllImport("shell32.dll", CharSet = CharSet.Auto)]
static extern bool ShellExecuteEx(ref SHELLEXECUTEINFO lpExecInfo);

[StructLayout(LayoutKind.Sequential, CharSet = CharSet.Auto)]
public struct SHELLEXECUTEINFO
{
    public int cbSize;
    public uint fMask;
    public IntPtr hwnd;
    [MarshalAs(UnmanagedType.LPTStr)]
    public string lpVerb;
    [MarshalAs(UnmanagedType.LPTStr)]
    public string lpFile;
    [MarshalAs(UnmanagedType.LPTStr)]
    public string lpParameters;
    [MarshalAs(UnmanagedType.LPTStr)]
    public string lpDirectory;
    public int nShow;
    public IntPtr hInstApp;
    public IntPtr lpIDList;
    [MarshalAs(UnmanagedType.LPTStr)]
    public string lpClass;
    public IntPtr hkeyClass;
    public uint dwHotKey;
    public IntPtr hIcon;
    public IntPtr hProcess;
}

private const int SW_SHOW = 5;
private const uint SEE_MASK_INVOKEIDLIST = 12;
public static bool ShowFileProperties(string Filename)
{
    SHELLEXECUTEINFO info = new SHELLEXECUTEINFO();
    info.cbSize = System.Runtime.InteropServices.Marshal.SizeOf(info);
    info.lpVerb = "properties";
    info.lpFile = Filename;
    info.nShow = SW_SHOW;
    info.fMask = SEE_MASK_INVOKEIDLIST;
    return ShellExecuteEx(ref info);        
}

// button click
private void button1_Click(object sender, EventArgs e)
{
    string path = @"C:\Users\test\Documents\test.text";
    ShowFileProperties(path);
}
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