[英]extraction of file from filepath
I need to extract file name without extension name. 我需要提取不带扩展名的文件名。
example. 例。
/home/si/text.txt /home/si/text.txt
/home/si/text.vx.txt /home/si/text.vx.txt
In the both case I should receive output text only. 在两种情况下,我都应该只接收输出文本。 I am not sure how many trailing extension file can have but I need to extract only file name.
我不确定尾随扩展文件可以有多少,但是我只需要提取文件名。 I have tried spliitext(filename)[0] but it gave me output text.vx rather than text
我已经尝试过spliitext(filename)[0],但是它给了我输出text.vx而不是text
This should work for your needs: 这应该可以满足您的需求:
from os.path import basename
print basename("/home/si/text.vx.txt").split('.')[0]
>>> text
I use split function after getting file name. 我在获取文件名后使用了分割功能。
filename.split('.')[0] filename.split('。')[0]
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