Like you, all of my research has come to the conclusion of "it depends". For
this particular product, we have an index of a million documents or so. And
each document can belong to many catalogs. Initially, it will be a small
number, but there could be up to 200 or so catalogs (probably much less). So,
for simplicity and speed of development, it probably makes the most sense to
just put the list of catalog IDs in the document. Sure, changes in the ACL will
cause a re-index of that product, but things don't change that often.
________________________________
From: Chris Hostetter <hossman_luc...@fucit.org>
To: solr-user@lucene.apache.org
Sent: Tue, February 23, 2010 3:40:39 PM
Subject: Re: filter result by catalog
: Yes I thought about both methods. The ACL method is easier, but has some
: scalability issues. We use the bitset method in another product, but
: there are some complexity and resource problems.
:
: This is a new project so I am revisiting the issue to see if anyone had any
better ideas.
The issues with something like this really depend on the specifics ... how
the rules of things "allowed to see" is defined, how often those rules are
changed, how many unique users you have, what kinds of inheritence the
rules need, etc...
for example: If your rules are as simple as
* "every doc is in exactly one catalog
* no doc ever changes catalog
* some catalogs require subscriber level
* the list of catalogs requireing subscriber level changes daily
...then it makes sense to index the catalog name as part of hte
documents, and have a simple two stage lookup -- pass in "subscriber"
or "not-subscriber" at runtime, and have a parser that looks at an
external list of subscriber catalogs and translates that into a filter at
runtime.
...if the "subscriber" catalogs never change, you can make it simpler and
index the subscriber/not-subscriber info directly as a field; if ocs
switch catalogs frequently, or are in multiple catalogs, or there are
more rules, or more complex hierarchical rules, then the implementation
becomes more involved.
but there's no single good answer.
-Hoss
