I have one Element (in this case Shipment No) with Multiple "textcode keys". You can see that by the Sum I do there (for columns right after the shipment no).
I have troubles to identify if one has more then one. You see in the picutre below i use in row
3 = IF((([Textcode key]='T98' )) ,'YES', 'NO') --> And it Works
4 = = IF((([Textcode key]='T99' )) ,'YES', 'NO') --> And it works
so now I tried in 1 and 2 to combine them without success.
First I tried 2 = IF((([Textcode key]='T99') and ([Textcode key]='T98' )) ,'YES', 'NO') --> doesn't work, always says no
and then 1 = MaxString( If(([Textcode key] =('T98') AND [Textcode key] =('T99')) ,Dual('YES', 2), Dual('no', 1)))
--> doesn't work, always says no
I expected to get "YES" in 1 or 2 but I always get "no".
Ideally I would delete 1 till 4 and have just one dimension that says yes if one or more Textcodes are found. All I need is some variaton of formular 1 or 2 that somehow reflects the combination of 3 AND 4.
ok, i could use this (see last column in screenhot)
=WildMatch ((aggr(concat( [Textcode key],'/') , [Shipment no])), '*T98/T99*') And use the IF clause looking for the 1.
It seems that Aggr(Concat) sort them Alphabetically so in this case where I look for the combination T98 and T99 it works. But if I needed to look for the combination T33 and T99 and in the dataset there would be a T50 aswell then the Aggr(Concat) would probably say T33/T50/T99 thus rendering the syntax (=WildMatch ((aggr(concat( [Textcode key],'/') , [Shipment no])), '*T33/T99*') useless