Im not finding it but is there something to increment by letters?

1a

1b

1c

kind of thing?

TIA

To increment what? And what should happen after "1z"?

Edited August 10, 200916 yr by Guest

Sorry for not being clear but the chances are we will never get to E let alone Z. Its for a process thats done usually once a day but in the past 2 years maybe 25 times we done it twice in a day and 3 times in one day is even more rare.

It is for a report and the letter will be appended to a date on record creation.

So it would actually look more like.

08102009A

08102009B

08112009A

08122009A

I am still not sure how exactly is the 'increment' supposed to happen. Assuming you know how to get it to increment by numbers, e.g.

08102009 1

08102009 2

08112009 1

08122009 2

you can convert the numbers to letters simply by:

Char ( 64 + number )

This is the solution I've some up with to solve my needs. I doubt this is the best solution and I know someone will probably poke giant holes in my calc but so far its working for me and I've not been able to break it.

Let ( 

[$Date=Year ( Get ( CurrentDate ) ) &

SerialIncrement ( "00" ; Month ( Get ( CurrentDate )  )  )& 

SerialIncrement ( "00"; Day ( Get ( CurrentDate )  ) ) ;



$count = If ( Count ( Manifest 2::zc_ManifestDate ) <1 or IsEmpty ( Count ( Manifest 2::zc_ManifestDate ) )  ; 1 ; Count ( Manifest 2::zc_ManifestDate )+1 )] ; 







Case ( 

$count =1 ; $Date & "A" ;

$count =2 ; $Date &"B" ;

$count =3 ; $Date &"C" ;

$count =4 ; $Date &"D" ;

$count =5 ; $Date &"E" ;

$count =6 ; $Date &"F" ;

$count =7 ; $Date &"G" ;

$count =8 ; $Date &"H" ;

$count =9 ; $Date &"I" ;

$count =10 ; $Date &"J" ;

$count =11 ; $Date &"K" ;

$count =12 ; $Date &"L" ;

$count =13 ; $Date &"M" ;

$count =14 ; $Date &"N" ;

 "Error" ) 



)