如何通过按钮
打开文件的属性对话框private void button_Click(object sender, EventArgs e)
{
string path = @"C:\Users\test\Documents\tes.text";
// how to open this propertie
}
谢谢。
例如,如果想要系统属性
Process.Start("sysdm.cpl");
但是如何获取文件路径的“属性”对话框?
答案 0 :(得分:46)
解决方案是:
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);
}
答案 1 :(得分:12)
调用Process.Start,传递包含文件名称的ProcessStartInfo,并将ProcessStartInfo.Verb设置为properties
。 (有关更多信息,请参阅非托管SHELLEXECUTEINFO结构的描述,这是ProcessStartInfo包含的内容,特别是lpVerb成员。)
答案 2 :(得分:7)
FileInfo类提供了各种文件属性:
FileInfo info = new FileInfo(path);
Console.WriteLine(info.CreationTime);
Console.WriteLine(info.Attributes);
...
答案 3 :(得分:0)
解决方案是使用ShellExecute ()
api。
如何使用C#调用此api: http://weblogs.asp.net/rchartier/442339
对于我而言,这在调试和发布模式下都没有CharSet
属性。