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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;
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Windows启动时如何运行我的应用程序?
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