How to access files inside a Python egg file?

问题:

This might be a weird requirement but it's what I've run into. I Googled but yield nothing.

I'm coding an application who's using a lot of constant attributes / values recorded in an XML file (they'll not change so a static file), things work fine until I generated an egg file for it.

When the logic reaches the XML accessing part, I got one complaint like this: /home/Workspace/my_proj/dist/mps-1.2.0_M2-py2.6.egg/mps/par/client/syntax/syntax.xml

Actually I've bundled the XML file in the path above but seems Python doesn't know how to access it.

The code to access the XML is as...

file_handler = open(path_to_the_file)
lines = file_handler.read().splitlines()

Any idea?

回答1:

egg files are zipfiles, so you must access "stuff" inside them with the zipfile module of the Python standard libraries, not with the built-in open function!

回答2:

If you want to access the contents inside the .egg file you can simply rename it and change extension from .egg to .zip and than unzip it. Which will create a folder and the contents will be same as they were when it was a .egg file

for example brewer2mpl-1.4.1-py3.6.egg
After Renaming brewer2mpl-1.4.1-py3.6.zip

Now if we open it, it'll get easily unzipped and the content will be put in a folder with same name in the same directory. (tested on macOS Sierra)

回答3:

Access file from inside egg file

Yes, It is possible to read the files from inside egg file.

Egg file: mps-1.2.0_M2-py2.6.egg structure for module level example:

In driverfile.py:

import xml.etree.ElementTree
import mps.par.client as syntaxpath
import os
path = os.path.dirname(syntaxpath.__file__)
element = xml.etree.ElementTree.parse(path+'\\syntax\\syntax.xml').getroot()
print(element)

Read xml file from inside an eggfile:

PYTHONPATH=mps-1.2.0_M2-py2.6.egg python driverfile.py

回答4:

just use unzip file.egg this should be enough.