Remove an element if some condition is met

[L30Bola]

Hi!!

I'd like some help on a yq command. Some context before:

I have a bunch of spec.yaml file that are structured like this:

namespace: test-namespace
releaseName: test-release
paused: false
environment: production
config:
  region: us-east-1

I wanted to first add a .config.image field on the production specs, which I could successfully with this:

for yaml in directory/**/spec.yaml; do
    yq -i '(select(.environment == "production") | .config.image) = "alpine:latest"' $yaml
done

With the above for-loop, only the production specs were updated. Now, if I want to delete the added field, I tried this:

for yaml in directory/**/spec.yaml; do
    yq -i '(select(.environment == "production") | del(.config.image))' $yaml
done

And also this:

for yaml in directory/**/spec.yaml; do
    yq -i '(select(.environment == "production")) | del(.config.image)' $yaml
done

The problem is that, even though on production spec.yaml the .config.image field were deleted, spec.yaml that have environment != production had it's entire content deleted. Is there a way to delete the field on production spec.yaml while ignoring non-production?

The first part to add a new value, I got from this other discussion: #617

EDIT:
Forgot to add the yq version:

$ yq --version
yq (https://github.com/mikefarah/yq/) version v4.44.3

[mikefarah]

Sure, the issue is that you are first filtering to only include files with environment = production, and then deleting nodes in those matches.

Instead do:

yq 'del(select(.environment == "production") | .config.image)' examples/data1.yaml

The filtering is inside the delete, so it will only delete those matching nodes, or leave the data alone.

[L30Bola]

Oh, very nice. It worked flawlessly!