Bimlas As I read this
- the first set/run is listed - The second run lists those in the first set but not those in the second set - Then the filter as a whole is the first set but *not* those "*not* in the second set" - This is a double negative so in effect saying is list the first set but *only* those in the second set. I now face the challenges of getting the set names from fields, and converting the set names into the list of members in those sets and passing those into your suggested filter. Thanks Tony On Friday, August 9, 2019 at 12:18:07 AM UTC+10, bimlas wrote: > > Not the best, but at least it works. [a b c] is the first set, [b c d] is > the second. > > [enlist[a b c]] -[enlist[a b c]!enlist[b c d]] > -- You received this message because you are subscribed to the Google Groups "TiddlyWiki" group. To unsubscribe from this group and stop receiving emails from it, send an email to tiddlywiki+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/tiddlywiki/014c443c-383a-44a4-9377-ea9125f4ad57%40googlegroups.com.
