在Windows启动时启动窗口 [英] Launch Window on Windows startup

问题描述

我想我的应用程序（一个WPF 窗口），在Windows启动时启动。我想尽了各种办法，但似乎没有人上班。我有我的代码写的做到这一点？

I want that my application (a WPF Window) is launched on Windows startup. I tried different solutions, but no one seems to work. What i have to write in my code to do this?

推荐答案

当你说你必须添加一个关键看你是正确的。注册表

You are correct when you say that you must add a key to the registry.

添加一键：

HKEY_CURRENT_USER\SOFTWARE\Microsoft\Windows\CurrentVersion\Run

如果您想启动应用程序

Or:

或\\Microsoft\Windows\CurrentVersion\Run

HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Run

如果你想要启动它为所有用户。

If you want to start it for all users.

例如，从当前用户的应用程序：

For example, starting the application for the current user:

var path = @"SOFTWARE\Microsoft\Windows\CurrentVersion\Run";
RegistryKey key = Registry.CurrentUser.OpenSubKey(path, true);
key.SetValue("MyApplication", Application.ExecutablePath.ToString());

只需要用

RegistryKey key = Registry.LocalMachine.OpenSubKey(path, true);

如果你想自动启动在Windows启动时所有用户的应用程序。

if you want to automatically start the application for all users on Windows startup.

只是删除注册表值，如果您不再希望自动启动的应用程序

Just remove the registry value if you no longer want to start the application automatically.

这样：

var path = @"SOFTWARE\Microsoft\Windows\CurrentVersion\Run";
RegistryKey key = Registry.CurrentUser.OpenSubKey(path, true);
key.DeleteValue("MyApplication", false);

此示例代码是为WinForms应用程序进行测试。如果您需要确定可执行文件路径为一个WPF应用程序，然后给一个尝试以下。

This sample code was tested for a WinForms app. If you need to determine the path to the executable for a WPF app, then give the following a try.

string path = System.Reflection.Assembly.GetExecutingAssembly().Location;

只需更换Application.ExecutablePath.ToString（）的路径可执行文件。

Just replace "Application.ExecutablePath.ToString()" with the path to your executable.

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