Windows启动时如何运行我的应用程序? [英] How to run my application when windows start up?

问题描述

我刚刚创建了一个弹出窗口.现在我想在窗口启动时自动运行它.请告诉我该怎么做?

Hi,

I just created a pop up window.Now i want to run it automatically when the windows start up.Kindly tell me how to do that?

推荐答案

将您的exe放入以下文件夹.它会在每个srartup上运行.为了进行测试，请重新启动Windows.

C:\ Documents and Settings \ Administrator \开始菜单\ Programs \ Startup

Put your exe in following folder. it will run at every srartup. For test, restart the windows.

C:\Documents and Settings\Administrator\Start Menu\Programs\Startup

为此，您需要创建Windows服务.
简单Windows服务示例 [如何在C#中开发Windows服务 [ ^ ]

For that you need to create windows service.
Simple Windows Service Sample[^]

How to develop Windows Service in C#[^]

此行将添加用于自动启动的注册表项:

This line will add a registry entry for autostart:

Registry.CurrentUser.CreateSubKey(@"Software\Microsoft\Windows\CurrentVersion\Run").SetValue("keyNameForRegistry", "PATH_TO_EXECUTABLE");

并且此行将删除自动启动:

And this line will remove the autostart:

Registry.CurrentUser.CreateSubKey(@"Software\Microsoft\Windows\CurrentVersion\Run").DeleteValue("keyNameForRegistry", false);

我相信这些应该有效.我现在没有时间测试它们.可能会有一些错别字.不要忘记添加，不确定是否两者都需要.

I believe these should work. I don''t have time to test them now. There might be some typos. Dont forget to add, not sure if the both are needed tho.

using Microsoft.Win32;
using System.Reflection;

这篇关于Windows启动时如何运行我的应用程序?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！