Provide example on skipping certain Python versions

[letmaik]

In my library I don't support Python 2.6 anymore, and I was wondering what the best way is to skip this version in the build-wheels.sh script, instead of the loop for PYBIN in /opt/python/*/bin.

[letmaik]

Sorry, just realized this should probably go to the https://github.com/pypa/python-manylinux-demo repo, right?

[rmcgibbo]

Cool. Is that something you would be able to contribute? It should just be a little bit of bash scripting.

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On May 23, 2016, at 4:08 AM, Maik Riechert notifications@github.com wrote:

Sorry, just realized this should probably go to the https://github.com/pypa/python-manylinux-demo repo, right?

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[letmaik]

I use this for my own library at the moment:

PYBINS=(
  "/opt/python/cp27-cp27m/bin"
  "/opt/python/cp27-cp27mu/bin"
  "/opt/python/cp33-cp33m/bin"
  "/opt/python/cp34-cp34m/bin"
  "/opt/python/cp35-cp35m/bin"
  )

# and later...
for PYBIN in ${PYBINS[@]}; do
  # ...
done

[njsmith]

There's a few different options depending on what exactly you want... for example /opt/python/cp{27,3}*/bin will give you all cpython 2.7 or 3.x variants, including new 3.x versions added in the future, but not pypy even if it's added in the future.

I guess it might be useful to have some example that explicitly uses a blacklist and gets all new variants by default - obviously projects can do what they want, but I think everyone's better off in the long run if we make it easy to add new interpreters and if projects that don't care that much then get automatically opted in.

Possibly the best is something like

for PYTHON in /opt/python/*/bin/python; do
    # only run body if this interpreter claims to be 2.7 or better
    if python -c 'import sys; sys.exit(not sys.version_info >= (2, 7))'; then
        # build here
    fi
done

(Or this could avoid an indent level if bash has a continue construct... anyone know if this exists?)

[ulope]

I've been wrapping the pip wheel command with if [[ ${PYBIN} != *"cp26"* ]]; then to exclude Python 2.6

[AvdN]

@njsmith Bash has a continue construct, it is called... continue ;-)

for x in $(seq 3) ; do if [ $x == "2" ] ; then continue; fi; echo $x; done