如何浏览目录中有空格的目录（开始菜单）？ [英] How to Explore a directory that the directory have Space (Start Menu)?

问题描述

String path = @C：\ProgramData \ Microoft \ Windows \Start Menu \Program;

Process.Start（path，* .lnk）;

解决方案

签名是： Process.Start（Filename，Arguments）;

您正尝试以* .lnk作为参数启动目录。我认为它不会起作用。您可以尝试这样的事情：

 Process.Start（@％windir％\ explorer.exe，@C：\ProgramData \ Microsoft \ Windows \\ \Start Menu \Program）; 

祝你好运！

你传递的是什么

 Process.Start（path，* .lnk）; 

在Process.Start（）方法的第二个参数命令行参数中。

在你的情况下，我认为你不需要传递命令行参数。所以留空。

 字符串 path =  @ < span class =code-string> C：\ProgramData \ Mycoftoft \ Windows \Start Menu \Program; 
 Process.Start（path， ）; 

String path = @"C:\ProgramData\Microsoft\Windows\Start Menu\Program";
Process.Start (path, "*.lnk");

解决方案

The signature is: Process.Start (Filename, Arguments);
You are trying to start a directory with "*.lnk" as parameter. I don't think it will work. You could try something like this:

Process.Start(@"%windir%\explorer.exe", @"C:\ProgramData\Microsoft\Windows\Start Menu\Program");

Good luck!

What are you passing in

Process.Start (path, "*.lnk");

in the Process.Start() method second parameter command line arguments.
In your case I think you need not to pass a command line argument. so leave it blank.

String path = @"C:\ProgramData\Microsoft\Windows\Start Menu\Program";
Process.Start (path, "");

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