windows 如何将多个文件名传递给上下文菜单 Shell 命令？
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How to Pass MULTIPLE filenames to a Context Menu Shell Command?
提问by WinWin
Passing a single filename to a context menu shell command is simple:
将单个文件名传递给上下文菜单 shell 命令很简单：
[HKEY_CLASSES_ROOT\*\shell\MyProgram\Command]
@="program.exe %1"
But if I select multiple files, program.exeis invoked for eachsuch selected file.
但是如果我选择多个文件，program.exe则会为每个这样选择的文件调用。
What I would like to do instead is invokeprogram.exeonly once, passing to it all the filenames currently selected.
我想做的是program.exe只调用一次，将当前选择的所有文件名传递给它。
How to do this?
这该怎么做？
采纳答案by Eternal Learner
回答by W4ldi
You can use Send Tofor this. It supports multiple files.
In case this website goes offline:
如果本网站离线：
Open shell:sendtowith Windows + Ror paste it into your explorer address bar. It should redirect you to:
打开shell:sendto用Windows + R或粘贴到你的资源管理器的地址栏。它应该将您重定向到：
C:\Users\<yourusername>\AppData\Roaming\Microsoft\Windows\SendTo
C:\Users\<yourusername>\AppData\Roaming\Microsoft\Windows\SendTo
Create a shortcut to your program in this folder and you should see it in your explorer right-click menu under Send to
在此文件夹中创建程序的快捷方式，您应该在资源管理器的右键单击菜单下看到它Send to
