How do I run a Perl one liner from a makefile?

Question

I know the perl one liner below is very simple, works and does a global substitution, A for a; but how do I run it in a makefile?

perl -pi -e "s/a/A/g" filename

I have tried (I now think the rest of the post is junk as the shell command does a command line expansion - NOT WHAT I WANT!) The question above still stands!

APP = $(shell perl -pi -e "s/a/A/g" filename)

with and without the following line

EXE = $(APP)

and I always get the following error

make: APP: Command not found

which I assume comes from the line that starts APP

Thanks

Answer 1

You should show the smallest possible Makefile which demonstrates your problem, and show how you are calling it. Assuming your Makefile looks something like this, I get the error message. Note that there is a tab character preceding the APP in the all: target.

APP = $(shell date)

all:
    APP

Perhaps you meant to do this instead:

APP = $(shell date)

all:
    $(APP)

I did not use your perl command because it does not run for me as-is. Do you really mean to use Perl's substitution operator? perl -pi -e "s/a/A/g"

Here is a link to GNU make documentation.

Answer 2

If you want to run perl as part of a target's action, you might use

$ cat Makefile
all:
        echo abc | perl -pe 's/a/A/g'
$ make
echo abc | perl -pe 's/a/A/g'
Abc

(Note that there's a TAB character before echo.)

Perl's -i option is for editing files in-place, but that will confuse make (unless perhaps you're writing a phony target). A more typical pattern is to make targets from sources. For example:

$ cat Makefile
all: bAr

bAr: bar.in
        perl -pe 's/a/A/g' bar.in > bAr
$ cat bar.in
bar
$ make
perl -pe 's/a/A/g' bar.in > bAr
$ cat bAr
bAr

If you let us know what you're trying to do, we'll be able to give you better, more helpful answers.