Tuesday, January 4, 2022

[SOLVED] Using shell in Makefile to find Ubuntu Version

Issue

I'm trying make Ubuntu version specific distros of my tool so I want to get the os name and version. I have the following code:

ifeq ($(OS),Windows_NT)
    OS_POD+=win
else
    UNAME_S := $(shell uname -s)
    ifeq ($(UNAME_S),Linux)
            OS_VERS := $(shell lsb_release -a 2>/dev/null | grep Description | awk '{ print $2 "-" $3 }')
            OS_POD=./dist/linux/$(OS_VERS)
    endif
    ifeq ($(UNAME_S),Darwin)
            OS_POD=./dist/mac
    endif
endif

I use the shell one-liner:

lsb_release -a 2>/dev/null | grep Description | awk '{ print $2 "-" $3 }'

Which properly returns Ubuntu-12.04.2 outside of the Makefile, but inside it it returns nothing. That is, the OS_VERS variable is just -.

How can I fix this?


Solution

In a Makefile, $ is special. Use $$ where you want the shell to find a dollar.

OS_VERS := $(shell lsb_release -a 2>/dev/null | grep Description | awk '{ print $$2 "-" $$3 }')


Answered By - choroba
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