Unzip a file using python caller in FME 2011

Question

The idea is to download a zipped file from a website and unzip it using FME 2011. I have attached a screenshot of FME workbench which is not working. Any help is greatly appreciated.

Thanks.

Sam

Answer 1

Your code seems incorrect. Try:

import pyfme, os, zipfile
z = zipfile.ZipFile("GoTransitDL.zip")
z.extractall()

And just as a warning (from API docs):

Warning Never extract archives from untrusted sources without prior inspection. It is possible that files are created outside of path, e.g. members that have absolute filenames starting with "/" or filenames with two dots "..".

The zip file can be made to extract files to anywhere on your hard drive, and z.extractall() will oblige. Very dangerous if you don't know the source.

Answer 2

I just did the exact same thing. You should check the documentation on the PythonCaller. You need to create a function or class as described in the help. Use a function to process a single feature, as in this case.

Example:

from zipfile import ZipFile

def fmeUnzip(fmeFeature):
    zipFile = fmeFeature.getAttribute("_ZIPFILE_NAME")
    outDir = fmeFeature.getAttribute("_OUTPUT_DIR")
    unzip(zipFile, outDir)

def unzip(zipFile, outDir):
    z = ZipFile(zipFile)
    z.extractall(outDir)

In the PythonCaller parameters, set 'Symbol to use' to fmeUnzip and copy the code to the source code parameter. You can alternatively save it in a .py file, see the PythonCaller help for how to do this.

Note: _ZIPFILE_NAME and _OUTPUT_DIR are attributes created by ParameterFetcher transformers, which create attributes based on the values of workspace parameters.