I have CSV files inside data/processed/
:
data
processed
file1.CSV
file2.CSV
file3.CSV
file4.CSV
file5.CSV
and I want to create an archive with all these 5 files:
data.ZIP
file1.CSV
file2.CSV
file3.CSV
file4.CSV
file5.CSV
I tried to execute this command:
!zip -r data.zip data/processed/
However, the ZIP file looks as follows:
data.zip
data
processed
file1.CSV
file2.CSV
file3.CSV
file4.CSV
file5.CSV
Use -j
to ignore all path information. From the man page:
-j
--junk-paths
Store just the name of a saved file (junk the path), and do not
store directory names. By default, zip will store the full path
(relative to the current directory).