Delete Kubernetes namespace only if it's empty?

I'd like destroy the namespace only if it is empty. Is there a command to do this?

No there is not command to do that. This behavior is by design.

I would suggest a different approach. You should have all your deployment yamls in version control system for all of the components including namespace. When you want to create use kubectl create -f deployment.yaml and when you want to delete use kubectl delete -f deployment.yaml

You can run kubectl get all --namespace YOUR_NAMESPACE and then depends on output call delete namespace

See Remove Empty Namespaces Operator, it can do exactly what you want.

Why? Because it's not so easy to iterate over resources in the namespace to decide if it's empty or not. After all, there are "default resources" like default service account and probably other stuff from you tooling/operators.

So these resources should be excluded from iteration. Bash scripting becomes too complicated this way. And one day I decided to implement it with Python.

There's not a simple command to check a namespace before delete, it requires some kubectl scripting or a kube API client.

From the github issue discussing get alls limitations liggit provides an example and adding some jq processing you can get a (slow) command that errors unless it successfully finds all resource types are empty (no items):

set -o pipefail
kubectl api-resources --verbs=list --namespaced -o name \
 | xargs -n 1 kubectl get --ignore-not-found -n staging -o json \
 | jq '.items[]| .kind + "/" + .metadata.name | error'

try this, better iterate over kube-api resources and this will give every resource list inside the namespace.

kubectl api-resources --verbs=list --namespaced -o name \
| xargs -n 1 kubectl get --show-kind --ignore-not-found -l <label>=<value> -n 
<namespace>

or another approch

kubectl api-resources --verbs=list --namespaced -o name | `
%{ kubectl get $_ --show-kind --ignore-not-found -l <label>=<value> -n 
<namespace> }