Makefile: using command output in conditional

Question

In a makefile, I want to use a shell command such as command1=[some program] [--flags], then use a condition to see whether or not the output contains a certain keyword using grep.

My code does not work:

var=$([command 1] | grep [keyword])
if [ -z ${var} ]; then \
[command 2, with variables in it]; \
[command 3, with variables in it]; \
fi

any idea why?

Put the commands you want to run in a shell script, and use [the `$(shell)` command](https://www.gnu.org/software/make/manual/make.html#Shell-Function) to run it and save its output in a variable. — Some programmer dude, May 24 '21 at 06:48
Thank you. I need some of the local variables from the script in the command. Is it still possible to do? — ntbd, May 24 '21 at 06:52

Renaud Pacalet · Answer 1 · 2021-05-24T07:09:48.950

What you show is not make syntax, it is shell syntax. So I will assume that it is part of a make recipe. Your problem is then that:

Each line of a make recipes is passed to a different shell. So, you cannot pass shell variables from one line to the next. Use one single shell command list and line continuation (; \) to pass your entire recipe to one single shell.
Make expands the recipes before passing them to the shell. Escape the make expansions by doubling the $ signs that you want to preserve ($${var}). Do the same for all shell variable expansions. You can also replace the $() command substitutions by the equivalent backtick notation `` instead of $$().
```
 target: prerequisite ...
     var=`[command 1] | grep [keyword]`; \
     if [ -z "$${var}" ]; then \
         [command 2, with variables in it]; \
         [command 3, with variables in it]; \
     fi
```

I can't upvote yet because I'm new, but this worked. Thank you very much. — ntbd, May 24 '21 at 08:39
Even if you can't upvote you can probably accept the answer, which will be useful to others trying to solve similar issues. — Renaud Pacalet, May 24 '21 at 09:14

Makefile: using command output in conditional

1 Answers1