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test: Eviction E2E #996
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test: Eviction E2E #996
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Addressed comment from #959 (comment) |
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It("should ensure no resources exist on evicted member cluster with taint", func() { | ||
taintClusters := buildClustersFromNames(taintClusterNames) | ||
for _, cluster := range taintClusters { |
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there is only one cluster in the taint list, why using a loop? Should we taint two clusters?
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buildClustersFromNames is a util function that takes in a list if cluster names and returns the list of E2E cluster objects. Hence we need to iterate through the list. In this file we only taint one cluster but we taint multiple clusters here for reference
fleet/test/e2e/taint_toleration_test.go
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taintClusterNames = []string{memberCluster1EastProdName, memberCluster2EastCanaryName} |
test/e2e/placement_eviction_test.go
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It("should still place resources on the all available member clusters", checkIfPlacedWorkResourcesOnAllMemberClusters) | ||
}) | ||
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var _ = Describe("ClusterResourcePlacement eviction of bound binding - PickAll CRP, PDB specified to protect resources on all clusters, eviction denied", Ordered, func() { |
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is there a pickFixed case? Is there a pickAll but PDB specified maxUnavailabe case?
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