SerialIncrement for Letters?

[grumbachr]

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

1a

1b

1c

kind of thing?

TIA

[comment]

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

Edited August 10, 2009 by Guest

[grumbachr]

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

[comment]

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 )

[grumbachr]

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" ) 



)