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php: pass $(MAKE) to run-php-tests.sh

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Instead of hardcoding "make" in run-php-tests.sh, pass the actual name
of make from the Makefile; the default is still "make", mostly to use
the script without having to set $MAKE.
This commit is contained in:
Pino Toscano
2016-03-18 15:35:51 +01:00
parent 39772344ad
commit 76a01fe41d
2 changed files with 4 additions and 2 deletions
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4
php/Makefile.am
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@@ -60,7 +60,9 @@ extension/config.h: extension/config.m4 ../config.status
test -f "$@" && touch -- $@
# Don't use --test because PHP always returns a success code.
TESTS_ENVIRONMENT = $(top_builddir)/run
TESTS_ENVIRONMENT = \
MAKE=$(MAKE) \
$(top_builddir)/run
TESTS = run-php-tests.sh

2
php/run-php-tests.sh
View File

@@ -44,4 +44,4 @@ printenv | grep -E '^(LIBGUESTFS|LIBVIRT|LIBVIRTD|VIRTLOCKD|LD|MALLOC)_' >> env
TESTS=$(echo tests/guestfs_*.phpt)
echo TESTS: $TESTS
make test TESTS="$TESTS" PHP_EXECUTABLE="$PWD/php-for-tests.sh" REPORT_EXIT_STATUS=1 TEST_TIMEOUT=300
${MAKE:-make} test TESTS="$TESTS" PHP_EXECUTABLE="$PWD/php-for-tests.sh" REPORT_EXIT_STATUS=1 TEST_TIMEOUT=300