I want to write a program to be added to the right-click menu and when I run the program from the right-click menu to create an index.php
file in current directory!
Exmaple:
I go to C:\wamp\www
Then I right-click somewhere and want to choose create index.php
This should create a new file called index.php
in the directory I clicked
Desired action:
A new file called index.php
wil be created at C:\wamp\www\index.php
More Details:
I want to make it look like "Text Document" in "New":
right-click => new => text document
when we click it windows create "New Text Document.txt" ... now i want when we click on my program windows create "index.php" file!
Registry path that this program will be operate within it:
HKEY_CLASSES_ROOT\Directory\Background\shell\
thanks to all ...
Have a look at this:
using System;
using System.IO;
namespace createIndex_php
{
class Program
{
static void Main(string[] args)
{
string path = string.Empty;
try
{
if (File.Exists(args[0])) // right clicked on a file in explorer
{
path = Path.GetDirectoryName(args[0]);
}
else
{
path = args[0];
}
}
catch (Exception ex)
{
Console.WriteLine(string.Format(@"Error while creating ""index.php"": {0}", ex.Message));
Console.ReadKey();
}
if (path != string.Empty)
{
path = Path.Combine(path, "index.php");
try
{
File.Create(path);
Console.WriteLine(@"""index.php"" created!");
Console.ReadKey();
}
catch (Exception ex)
{
Console.WriteLine(string.Format(@"Error while creating ""index.php"": {0}", ex.Message));
Console.ReadKey();
}
}
}
}
}
This will do the creation of the index.php
file.
Anyway, you have to subscribe your app to the right click menu using regedit
and go to following key:
HKEY_CLASSES_ROOT\*\shellex\ContextMenuHandlers
Please see the following links for that: