Does this do what you need? Aa Bb +[nsort[modified]limit[1]]
Or try something like this in 5.1.24 pre-release: Aa Bb :sort:date[get[modified]] +[limit[1]] See https://tiddlywiki.com/prerelease/#Sort%20Filter%20Run%20Prefix On Wednesday, May 5, 2021 at 7:41:37 AM UTC+2 Mohammad wrote: > Back to winter 2019, when we were working on the Commander plugin we > needed a comparison between two tiddlers to see which is greater. See [1] > The best solution we found (by BTC) at that time was as below [2] (still > works and is part of Commander) > > > <$set name="tidBMod" value={{{ [[Bb]get[modified]] }}}> > <$set name="tidAMod" value={{{ [[Aa]get[modified]] }}}> > <$list filter="[<tidAMod>] [<tidBMod>] > +[nsort[]last[1]removesuffix<tidAMod>]"> > Yes, Aa is newer > </$list> > <$list filter="[<tidAMod>] [<tidBMod>] > +[nsort[]last[1]removesuffix<tidBMod>]"> > No, Aa is older > </$list> > </$set> > </$set> > > > With the advent of many new filter operator, I want to know if there is a > shorter and more semantic solution can be found today Apr 2021 > > I thought about the below solution and it works in TW 5.1.23 > > <$vars dtb={{{[[Bb]get[modified]]}}}> > > <$list filter="[[Aa]get[modified]compare:integer:gt<dtb>then[Aa]else[Bb]]"> > <<currentTiddler>> is newer! > </$list> > > </$vars> > > > But I am wondering if we could use :filter or some other operator to drop > the $vars and dtb and perform the comparison in single $list > Note Aa and Bb are two tiddlers > > > > Best wishes > Mohammad > > [1] https://groups.google.com/g/tiddlywiki/c/9QQvkJkLogA/m/3AGSBBJKBgAJ > [2] https://groups.google.com/g/tiddlywiki/c/9QQvkJkLogA/m/UbIgaSFSBgAJ > > -- 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 [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/tiddlywiki/918a183b-82d2-411e-8896-06315800770bn%40googlegroups.com.
