找到一个文件与具体的扩展在任何地方的系统

0

的问题

我需要代码一个测试一个函数,我有义Python3.9. 作为功能是应该收到文件的一个特定的扩展及其路径,并随后将其装载在蟒蛇,我的测试应该寻找任何可用文件的扩展,并试图负载正确的。 怎么我可以代码这个吗?

file function path python
2021-11-23 21:55:23
1

最好的答案

0

也许这可以帮助:

import os
file_path = '/' # Location of files
file_extension = 'xlsx'
for i in os.listdir(file_path):
    if file_extension == i.split('.')[-1]:
        print(i)  # The task you want to perform with the specific files

在这里, file_path 是路径的文件列表, file_extension 是特定的扩展,我们正在寻找。

2021-11-24 00:44:48

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