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fix: use policyName key to get the policy name (#4114)

In case of namespace policy `ur.spec.policy`
contains namespace/policy-name combinations, hence
can't be used to set the policy name label.

Signed-off-by: prateekpandey14 <prateek.pandey@nirmata.com>

Co-authored-by: shuting <shuting@nirmata.com>
This commit is contained in:
Prateek Pandey 2022-06-16 16:53:37 +05:30 committed by GitHub
parent bc1b051b90
commit 02b806deee
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@ -71,10 +71,11 @@ func (g *generator) applyResource(policyName string, urSpec kyvernov1beta1.Updat
func (g *generator) tryApplyResource(policyName string, urSpec kyvernov1beta1.UpdateRequestSpec) error { func (g *generator) tryApplyResource(policyName string, urSpec kyvernov1beta1.UpdateRequestSpec) error {
l := logger.WithValues("ruleType", urSpec.Type, "kind", urSpec.Resource.Kind, "name", urSpec.Resource.Name, "namespace", urSpec.Resource.Namespace) l := logger.WithValues("ruleType", urSpec.Type, "kind", urSpec.Resource.Kind, "name", urSpec.Resource.Name, "namespace", urSpec.Resource.Namespace)
var queryLabels labels.Set var queryLabels labels.Set
if urSpec.Type == kyvernov1beta1.Mutate { if urSpec.Type == kyvernov1beta1.Mutate {
queryLabels = common.MutateLabelsSet(urSpec.Policy, urSpec.Resource) queryLabels = common.MutateLabelsSet(policyName, urSpec.Resource)
} else if urSpec.Type == kyvernov1beta1.Generate { } else if urSpec.Type == kyvernov1beta1.Generate {
queryLabels = common.GenerateLabelsSet(urSpec.Policy, urSpec.Resource) queryLabels = common.GenerateLabelsSet(policyName, urSpec.Resource)
} }
urList, err := g.urLister.List(labels.SelectorFromSet(queryLabels)) urList, err := g.urLister.List(labels.SelectorFromSet(queryLabels))
if err != nil { if err != nil {