S S ,
Yes exactly. I also tested
[enlist<list-of-tiddlers>] -[enlist{!!list}]
where <list-of-tiddlers> was a macro
I am keen to extend this to unions, not in either set etc...
Regards
Tony
On Tuesday, February 26, 2019 at 9:51:40 AM UTC+11, S. S. wrote:
>
> Actually, I jumped the gun there, as I was testing using the *enlist*
> operator when I read your post:
>
> This is what would give : {!!list} *contains *<myfilter>
>
> {{!!list}} = [[C A]] [[A C]] [[B B]] x
>
> field CONTAINS list:
> <$set name="check-list" filter="[[C A]] [[B B]] [[A C]]">
> <$list filter="[enlist<check-list>] -[enlist{!!list}]" emptyMessage=
> "MATCH">
> NO-MATCH
> </$list>
> </$set>
>
>
>
> On Tuesday, February 26, 2019 at 5:31:58 AM UTC+7, S. S. wrote:
>>
>> Tony, your suggestion is close. What it tells me is if <myfilter>
>> *contains* {!!list} - which is actually also useful to me!
>>
>> On Tuesday, February 26, 2019 at 4:42:31 AM UTC+7, TonyM wrote:
>>>
>>> S s
>>>
>>> A possible trick, in a filter generate the titles in one and remove the
>>> title found in the other. Emptyvalue or emptymessage will mean a match,
>>> including when both are empty.
>>>
>>> If not empty The result will be tiddlers not found in either list.
>>>
>>> To simplify it use the subfilter operator for both sets of tiddler and
>>> try -[subfilter[name]
>>>
>>> Eg untested
>>> "[subfilter<myfilter>] -[subfilter{!!list}]"
>>>
>>> Regards
>>> Tony
>>>
>>>
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