I create a shared content rule for multiple folders, but I couldn't find a way in the action executor to get the folder in which the rule was executed. Any idea?
i guess the rule should give you the nodeRef of the file and from there you can get the parent (if it is only 1)https://community.alfresco.com/docs/DOC-5359-noderef-cookbook
ChildAssociationRef childAssociationRef = nodeService.getPrimaryParent(nodeRef);NodeRef parent = childAssociationRef.getParentRef();
Yes, but this is a shared rule between multiple folders, it has many parents.
well then i don't understand - what are you trying to achieve? and how is your rule configured (screenshot)?
I define the rule under a static folder, then I link it to another one. this is an example of parent relations between the rule and the rule folders.
please explain what you are trying to achieve and how the rule is triggered
The rule is triggered when an element in the folder is created, deleted or updated. I'm trying to get the NodeRef of this folder in the actionExecuter.(the rule is linked to the folder not created under its ruleFolder.
The actionedUponNodeRef should be the noderef of the newly created File/folder and from there you can geht the Patent - no?
The actionedNodeRef is not always a direct child of the folder, it can be created in a subfolder.
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