You have standard module zipfile
to create ZIP, and glob.glob()
or os.listdir()
or os.walk()
to get filenames in folder.
EDIT: should works (I works for me on Linux)
import os
import zipfile
folder = 'C:\\My_Program\\zip_files'
for filename in os.listdir(folder):
if filename.endswith('.xlsx'):
name_without_extension = filename[:-5] # string `.xlsx` has 5 chars
xlsx_path = os.path.join(folder, filename)
zip_path = os.path.join(folder, name_without_extension + '.zip')
zip_file = zipfile.ZipFile(zip_path, 'w')
# use `filename` (without folder name) as name inside archive
# and it will not create folders inside archive
zip_file.write(xlsx_path, filename)
zip_file.close()
EDIT: the same with glob
import os
import glob
import zipfile
folder = 'C:\\My_Program\\zip_files'
for file_path in glob.glob(folder+'\\*.xlsx'):
filename = os.path.basename(file_path)
print(filename)
name_without_extension = filename[:-5]
print(name_without_extension)
xlsx_path = os.path.join(folder, filename)
zip_path = os.path.join(folder, name_without_extension + '.zip')
zip_file = zipfile.ZipFile(zip_path, 'w')
# use `filename` (without folder name) as name inside archive
# and it will not create folders inside archive
zip_file.write(xlsx_path, filename)
zip_file.close()