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Hello how can I get the file name with os.path lib? For example:

C:\Users\filippo\Desktop\K.java

I want the K without the extension file

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6

I suggest you use the splitext and basename functions from os.path

K, ext = os.path.splitext(os.path.basename(my_path))

See the docs here.

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You can achieve this using:

import os filename = r"C:\Users\filippo\Desktop\K.java" print os.path.splitext(filename)[0] > C:\Users\filippo\Desktop\K print os.path.splitext(filename)[1] > .java K, ext = os.path.splitext(os.path.basename(filename)) print K print ext > K > .java
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