Output of bash command substitution? #26547
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Hi, In our project, we have a Makefile, with a step to build our docker container. The real version is a bit more complex than the example below, but it captures the gist of what we’re doing.
Which works great. The problem is that since we need to capture output of the make recipe, we need to call it using a bash command substitution The exact same setup works fine locally on our machines, but the stdout from the inner Does anyone have suggestions on how we could continue capturing the output, while also getting the output of the Thanks! |
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Replies: 4 comments
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How about writing |
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To be honest I’d like to avoid creating a bunch of artifacts from my Makefile since we run them locally as well. Like I said, this works flawless locally since it’s not Make that’s hiding the output, it seems to be GitHub Actions Running |
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Then copy the output to a file in the Actions workflow before processing it, something like this:
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I decided to go another way. Since the generated container image name (and tag) is idempotent for the same commit, I decided to add a |
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I decided to go another way. Since the generated container image name (and tag) is idempotent for the same commit, I decided to add a
docker-info
recipe in my Makefile. I simply call that to get the name