Thank you for your answer, I appreciate your help.
However, your suggestion does not seem to affect the underlying problem: 
I'm unable to get the number of tiddlers that pass the filter of second 
list.
I'll try to elaborate what I meant in my original post.

For example let's say the outer list alone would output 100 tiddlers. If I 
added count[] to this filter to get the number of tiddlers this outer list 
outputs I will see the number 100 instead of 100 individual titles.
Then the inner list filters out 30 tiddlers that do not satisfy the 
additional condition. I see 70 tiddler titles that pass the outer AND inner 
filter, but I'm unable to count them (except by hand).
Adding count[] to the inner list would just yield a mix of 30 zeroes (0 0 0 
0...for each time the currently evaluated tiddler does not pass the inner 
filter) and 70 ones (1 1 1 1...whenever it does).

Now either I'd need some way to have a variable that would increase each 
time a tiddler passes the inner filter, or I'd need to have it all in a 
single-level list (which with current setup doesn't work exactly for the 
reason Eric describes). Or there could be a completely different approach 
that I'm unable to see.

0
maanantai 13. syyskuuta 2021 klo 18.02.58 UTC+3 Eric Shulman kirjoitti:

> On Monday, September 13, 2021 at 6:51:14 AM UTC-7 [email protected] wrote:
>
>> Aside: we could merge the two <$list> filters together 
>
>
> That wouldn't work in this case, as the inner filter uses `{!!frequency}`, 
> which depends upon the outer filter to set the `currentTiddler` value to 
> each tiddler that `has[frequency]`
> If the two filters were merged, then `{!!frequency}` would refer to the 
> tiddler that contains the `$list` widget, rather than each tiddler that 
> `has[frequency]`
>
> -e
>

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