Fallback on PYTHON path in Makefile

[jjwatt]

SUMMARY

If I understand correctly, this change should make '$(PYTHON)' work how we want it to everywhere. Before this change, on develpers' machines that don't have a 'python3.9' in their path, make would fail. With this change, we will prefer python3.9 if it's available, but we'll take python3 or python otherwise.

I think setting this might be an artifact from having several minor versions of Python 3 available. It may not even still be relevant. Let me know if I'm misunderstanding this or if there's somewhere where this will fail. My main goal is to avoid make failing because a developer does not have a bin named python3.9 in their path (or the equiv symlink or w/e). Like, maybe I have python3 and python3.11 and then I have to set PYTHON myself for running make. I think we can find something that works for these cases.

I really think this does what we mean for all cases: make from dev machine, make from container, make from scripts. It will even run the venv path one if started from venv.

ISSUE TYPE

• Bug, Docs Fix or other nominal change

COMPONENT NAME

• API
• UI
• Collection
• CLI
• Docs
• Other

AWX VERSION

ADDITIONAL INFORMATION

[shanemcd]

🍰

[jjwatt]

Hah. This is a fun one. So, whatever shell (version of sh) that GitHub runs did not like my trick with command -v python3 python3.9 It works it a lot of shells because it will return the first full path that actually exists. However, apparently the shell that runs from GH Actions -> ci.yml -> make -> ansible -> build.yml -> make print-VERSION is one that does not work that way! Instead it returns blank/empty if the first command can't be found :). So, that was causing the Permission Denied error I was getting in the awx-operator CI.

I fixed it by using a for loop in the shell macro instead of relying on my assumed command -v behavior. I still use command -v but in a loop, and it won't error out on missing commands.

I will clean this up and squash.

[TheRealHaoLiu]

related to #13737