Remove sandbox and subclusters from vdb at the same time #959
Conversation
roypaulin
requested review from
cchen-vertica,
fenic-fawkes,
HaoYang0000,
qindotguan and
LiboYu2
as code owners
| vdb.Spec.Subclusters = vdb.Spec.Subclusters[1:] | ||
| vdb.Spec.Subclusters = vdb.Spec.Subclusters[:2] | ||
| scNames = append(scNames[0:1], scNames[3]) | ||
| verifySubclusters(ctx, vdb, scNames, []int32{0, 0}, sbName, FindNotInVdbAcrossSandboxes) |
There was a problem hiding this comment.
Why the size is set to {0, 0}?
There was a problem hiding this comment.
Because when the operator constructs subclusters no longer in vdb from their statefulsets, it gives them size 0 as they will soon be removed.
api/v1/verticadb_webhook.go
Outdated
| // find subclusters that are sandboxed in old vdb but removed in new vdb | ||
| // checkSandboxSubclustersRemoved checks subclusters that are sandboxed in old vdb but removed in new vdb | ||
| func (v *VerticaDB) checkSandboxSubclustersRemoved(allErrs field.ErrorList, oldObj *VerticaDB, oldScIndexMap map[string]int, | ||
| oldScMap, newScMap map[string]*Subcluster, path *field.Path) field.ErrorList { | ||
| oldScInSandbox := oldObj.GenSubclusterSandboxMap() |
There was a problem hiding this comment.
We should modify this to: newScInSandbox := v.GenSubclusterSandboxMap(). Then in line 1647, verify the removed sc isn't in newScInSandbox. The check in line 1650(if v.GetSandbox(sb) == nil) is not accurate since the sc could be in a new sandbox of new vdb.
There was a problem hiding this comment.
I see, I always add the impression that moving a subcluster from one sandbox to another would not work. Then I need to change my definition of zombie subcluster as it will also cover the case where sandbox still exists.
This fixes the issue that happens when you remove a sandbox and all its subclusters at the same time from the vdb, the pods will be down but will never get deleted.
Now, as soon as unsandbox is done the operator will find the statefulset make it rejoin the main cluster where it will be removed.