CODEWITHSUNDEEP
bashshellmakefile
aahnik
aahnik

Reputation: 1861

This bash command is not running correctly inside Makefile

Inside Makefile I have like this:

release:
    version=$$(poetry version | cut -f2 -d " ")
    echo "release $$version"

If I run,the statements from my terminal they run without any problem.

> version=$(poetry version | cut -f2 -d " ")
> echo "release $version"
release 0.0.2

But if I run,

> make release                                                                                                                                  
version=$(poetry version | cut -f2 -d " ")
echo release $version
release

You see in the output, beside release the version no. is not shown.

Upvotes: 2

Views: 1110

Answers (1)

Barmar
Barmar

Reputation: 781004

Each command in a makefile recipe is executed in its own shell process. So the variable assignment takes place on one shell process, which then exits and its variables are discarded. The echo command executes in a new process that doesn't have that variable.

You need to escape the newline and use a ; command delimiter to run the commands in the same process.

release:
    version=$$(poetry version | cut -f2 -d " "); \
    echo "release $$version"

Upvotes: 5

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