'How do I convert a ZipFile object to an object supporting the buffer API?
I have a ZipFile object that I need to convert to an object that will work with the buffer api. Context is that I am trying to use an API that says it takes a file with the type string($binary)
. How do I do this? I know this is completely wrong, but here is my code:
def create_extension_zip_file(self, path_to_extension_directory, directory_name):
zipObj = ZipFile("static_extension.zip", "w")
with zipObj:
# Iterate over all the files in directory
for folderName, subfolders, filenames in os.walk(path_to_extension_directory):
for filename in filenames:
# create complete filepath of file in directory
filePath = os.path.join(folderName, filename)
with open(filename, 'rb') as file_data:
bytes_content = file_data.read()
# Add file to zip
zipObj.write(bytes_content, basename(filePath))
return zipObj
Solution 1:[1]
Or if the API expects a file-like object, you could pass a BytesIO instance when creating the zipfile and pass that to the API
import io
def create_extension_zip_file(self, path_to_extension_directory, directory_name):
buf = io.BytesIO()
zipObj = ZipFile(buf, "w")
with zipObj:
# Iterate over all the files in directory
for folderName, subfolders, filenames in os.walk(path_to_extension_directory):
for filename in filenames:
# create complete filepath of file in directory
filePath = os.path.join(folderName, filename)
with open(filename, 'rb') as file_data:
bytes_content = file_data.read()
# Add file to zip
zipObj.write(bytes_content, basename(filePath))
# Rewind the buffer's file pointer (may not be necessary)
buf.seek(0)
return buf
If the API expects a bytes instance, you could just open the zip file in binary mode after it has been written, and pass the bytes .
with open('static_extension.zip', 'rb') as f:
bytes_ = f.read()
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | snakecharmerb |