On OS X, link libpython normally, ignoring the "framework" framework.
As of Xcode 5.0, Apple isn't including the Python framework as part of the
SDK-level files, which means that linking to it might fail depending on
whether Xcode thinks you've selected a specific SDK version. According to
their Tech Note 2328, they've basically deprecated the framework method of
linking to libpython and are telling people to link to the shared library
normally. (I'm pretty sure this is in direct contradiction to the advice
they were giving a few years ago, but whatever.) Testing says that this
approach works fine at least as far back as OS X 10.4.11, so let's just
rip out the framework special case entirely. We do still need a special
case to decide that OS X provides a shared library at all, unfortunately
(I wonder why the distutils check doesn't work ...). But this is still
less of a special case than before, so it's fine.
Back-patch to all supported branches, since we'll doubtless be hearing
about this more as more people update to recent Xcode.
Branch
------
REL9_3_STABLE
Details
-------
http://git.postgresql.org/pg/commitdiff/4f5f4da79ee805681f1f23a107bc905d647f12bc
Modified Files
--------------
config/python.m4 | 11 ++---------
configure | 11 ++---------
src/pl/plpython/Makefile | 9 ++++++---
3 files changed, 10 insertions(+), 21 deletions(-)
