os.system调用一个位于名称中包含空格的dir中的exe [英] os.system to invoke an exe which lies in a dir whose name contains whitespace

问题描述

我的代码如下:

file = 'C:\\Exe\\First Version\\filename.exe'
os.system(file)

运行此程序时，出现Windows错误:can't find the file specified.

When I run this program, a Windows error is raised: can't find the file specified.

我发现问题出在第一版"中间的空白.我如何找到解决该问题的方法?

I found out the problem has to with the whitespace in the middle of "First Version". How could I find a way to circumvent the problem?

P.S .:如果将变量文件"作为参数传递给另一个函数怎么办?

P.S.: what if the variable 'file' is being passed as an argument into another function?

推荐答案

在路径周围加上引号将起作用:

Putting quotes around the path will work:

file = 'C:\\Exe\\First Version\\filename.exe'
os.system('"' + file + '"')

，但是更好的解决方案是改用subprocess模块:

but a better solution is to use the subprocess module instead:

import subprocess
file = 'C:\\Exe\\First Version\\filename.exe'
subprocess.call([file])

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