打开另一个程序配置文件C＃

Question

I have program A windows application and program B another windows application. In program BI need to open program A config file. (App.config) I want to open that in a note pad. How can I do this?

Answer 1

您可以使用Process.Start方法启动notepad.exe并指定文件的路径：

Process.Start("notepad.exe", "c:\path_to_a\app.config");

Answer 2

Very Simple ,

string filename = "C:\Users\Documents\Visual Studio 2010\Projects\myproject\app.config";

Start new process

Process.Start("notepad.exe",filename);