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批处理:在特定程序中打开特定文件?

[英]Batch: Open a specific file in a specific program?

当记事本是 .txt 文件的默认程序时,我如何告诉 Windows 在写字板中打开 C:\\test\\test.txt?

The accepted answer didn't work for me.接受的答案对我不起作用。 I am not sure if it was because of the program I was trying to run, or because the path had spaces in (even though I wrapped it in quotes), or something else.我不确定是因为我试图运行的程序,还是因为路径中有空格(即使我用引号将其包裹起来),还是其他原因。

Anyway, I was able to do it be adding an empty string after the start command.无论如何,我能够通过在start命令之后添加一个空字符串来做到这一点。

For example:例如:

start "" "C:\My Programs\myprogram.exe" "C:\My Files\myfile.txt"

您可以将直接路径添加到可执行文件,如

start C:\Windows\System32\write.exe [FILE]

尝试:

start wordpad c:\test\test.txt

There is no need to use start when using write.exe , so simply:使用write.exe时不需要使用start ,所以简单:

write [FILE(S)]

write.exe is located in C:\\Windows\\System32 . write.exe位于C:\\Windows\\System32

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