如何在Windows应用程序（.net）中启动

Question

i created a small tool in windows application using (.net language C#). i created setup for my tool and also when we click on minimize button it will be in system tray. My requirement is i want to place my tool in start up (start>All Programs>Start up) when i start my system automatically my tool is open this is my requirement please help me thank you.

Answer 1

转到解决方案资源管理器并选择您创建的设置文件夹，选择任何文件（项目设置），您会看到两个分割的窗口，一个是目标计算机上的文件系统，右键单击空白空间，您可以得到“添加特殊文件夹>>用户启动”文件夹”，您可以得到一个文件夹，然后右键单击并添加所有项目输出。

Answer 2

Just put the EXE you want to run to the correct startup folder. Username is the logged on user. Run cmd, then type echo %username% to find your username.

Windows Vista

c:\Users\[username]\AppData\Roaming\Microsoft\Windows\Start Menu\Programs\Startup

Windows XP/2000

c:\Documents and Settings\[username]\Start Menu\Programs\Startup

Windows NT 4

c:\Winnt\Profiles\[username]\Start Menu\Programs\Startup

Windows 95/98/ME

c:\WINDOWS\Start Menu\Programs\Startup

Answer 3

The cleaner way is to copy the exe to the following location

Environment.GetFolderPath(System.Environment.SpecialFolder.Startup)

The only reason I recommend using this Method instead of hardcoding the value , is because the paths can change and trying to pre-empt ever every users move is a waste of time.Should just mention that the method above still returns the same folders as the first answer