How do I validate a GNU Makefile parameter exists in a string - bash

Question

I am trying to get this rule to work. Its totally blowing up my whole Makefile.
It should validate that the value for version the user passed exists in the PYVERSION variable.
The actual behavior is mixed because I've tried so many different things. I'm no GNU expert so I've tried ${version} and $(version) and $version and the same things for PYVERSION. The current version I've posted here always goes to the else block, not matter the version input.
PYVERSIONS := "3.5 3.6 3.7 3.8"
.PHONY: venv
venv:
if test $(findstring ${version}, $(PYVERSIONS)); then
/Library/Frameworks/Python.framework/Versions/${version}/bin/python3 -m venv venv
else
$(error Bad python version given (${version}) project only supports ${PYVERSIONS})
fi;
calling it like make venv version=3.5 should successfully execute the if block and create the venv.
calling it like make venv version=2.7 should raise the error message in the else block.
I was trying to follow this post
Update
With help from #Beta, this is the final solution I got to work.
PYVERSIONS := "3.6 3.7 3.8"
PYPATH := /Library/Frameworks/Python.framework/Versions
.PHONY: venv
venv:
#if test $(findstring ${version}, $(PYVERSIONS)) ; \
then \
echo "Creating virtual environment for python ${version}"; \
$(PYPATH)/${version}/bin/python3 -m venv venv; \
else \
echo "Unsupported python version (${version}). Project supports $(PYVERSIONS)"; \
fi

Answer 1

Actually disagree with #Beta here. Since the answer is known before you even start executing the recipe, $(error …) is the clear way to go.
Now, make stores the entire recipe as a single recursively expanded variable. When make decides to build venv (for instance, you type make venv), it will expand the entire recipe, before passing each resulting line one-by-one to fresh instances of the shell.
Thing is, when make expands your recipe,
it always expands $(error Bad python version given …), and make stops even before calling the shell.
How do we get around this?
PYVERSIONS := 3.6 3.7 3.8
PYPATH := /Library/Frameworks/Python.framework/Versions
pyversion = $(or $(filter ${version},${PYVERSION}),$(error $$version [${version}] must be exactly one of ${PYVERSIONS}))
.PHONY: venv
venv:
echo "Creating virtual environment for python ${version}"
$(PYPATH)/${pyversion}/bin/python3 -m venv venv
So,
You type make venv
Make expands the recipe
Make expands pyversion
Make expands $(filter ${version},3.6 3.7 3.8)
If you have not set version on the command-line or environment to one of the three blessed strings,
make stops there and then with a helpful message NICE
OTOH if it has been set appropritely, make is happy and starts executing the recipe
Other nice characteristics:
No shell syntax
The exit code of python3 is returned to make (a big hole in your original recipe imho).
Things always look better in make rather than the shell.

Answer 2

You are mixing shell syntax with Make syntax.
If you want to handle this with a shell conditional, you must put the whole thing on one line in the recipe (since each line executes in its own subshell):
venv:
if test $(findstring ${version}, $(PYVERSIONS)) ; then echo $(version)/bin/python3 venv; else echo bad version $(version); fi
or you can wrap the line by means of backslashes:
venv:
if test $(findstring ${version}, $(PYVERSIONS)) ; \
then \
echo $(version)/bin/python3 venv; \
else \
echo bad version $(version); \
fi
(Note that there is only one TAB, in front of if.)
If you want to use a Make conditional within the rule:
venv:
ifeq ($(findstring ${version}, $(PYVERSIONS)),)
echo bad version $(version);
else
echo echo $(version)/bin/python3 venv;
endif
If you want to use the Maker error command, you have a problem. Make will evaluate the conditional before executing any rule, so if you use error, then if no valid version number is given, Make will throw the error even if venv was not the target. It is possible to use error, but it's a pain, so you must first decide whether the effect is worth the price.