What is the proper syntax for getting a Makefile to print the output directory of one of its output zip files?
- by 9exceptionThrower9
I'm trying to edit an Android Makefile in the hopes of getting it to print out the directory (path) location of one the ZIP files it creates. Ideally, since the build process is long and does many things, I would like for it print out the pathway to the ZIP file to a text file in a different directory I can access later:
Pseudo-code idea:
# print the desired pathway to output file
print(getDirectoryOf(variable-name.zip)) > ~/Desktop/location_of_file.txt
The Makefile snippet where I would like to insert this new bit of code is shown below. I am interested in finding the directory of $(name).zip (that is specific file I want to locate):
# -----------------------------------------------------------------
# A zip of the directories that map to the target filesystem.
# This zip can be used to create an OTA package or filesystem image
# as a post-build step.
#
name := $(TARGET_PRODUCT)
ifeq ($(TARGET_BUILD_TYPE),debug)
name := $(name)_debug
endif
name := $(name)-target_files-$(FILE_NAME_TAG)
intermediates := $(call intermediates-dir-for,PACKAGING,target_files)
BUILT_TARGET_FILES_PACKAGE := $(intermediates)/$(name).zip
$(BUILT_TARGET_FILES_PACKAGE): intermediates := $(intermediates)
$(BUILT_TARGET_FILES_PACKAGE): \
zip_root := $(intermediates)/$(name)
# $(1): Directory to copy
# $(2): Location to copy it to
# The "ls -A" is to prevent "acp s/* d" from failing if s is empty.
define package_files-copy-root
if [ -d "$(strip $(1))" -a "$$(ls -A $(1))" ]; then \
mkdir -p $(2) && \
$(ACP) -rd $(strip $(1))/* $(2); \
fi
endef
