Why not make a dashboard that has a Count for the Day on it. If you
have
really simple people this might work?
On Thu, Oct 11, 2018 at 9:54 AM Frank Cazabon <frank.caza...@gmail.com>
wrote:
Thanks, Obscuring the number is not a problem, but I think they want
to
be able to read the number off the ticket at the end of the day in
sequential order and I don't see how I can come up with a simple way
to
do this.
Frank.
Frank Cazabon
On 10/10/2018 03:48 PM, Richard Kaye wrote:
One way is to come up with a 10 letter "word" with each letter
representing a digit. You then convert the numeric sequence to alpha
characters. And to avoid the pattern being too easily spotted, add
one or
more random characters to the final string. It's not the most robust
encryption scheme in the world but it should obscure it enough.
Here's a
bit of VFP code to do this using the non-dictionary phrase MONKEYSHIP
as
the rebus. Pass it the number to encode and optionally .t. as the
second
parameter to append a single random character at the end.
**************************************
* Program: MONKEYSHIP.PRG
* Date: 05/07/2007 03:26 PM
* VFP Version: Visual FoxPro 09.00.0000.7423 for Windows
* Notes:
**************************************
FUNCTION MonkeyShip(m.tnStringToEncode AS Number, m.tlRandom AS
Boolean)
LOCAL m.lcRetval AS Character
m.lcStringToEncode=ALLTRIM(TRANSFORM(m.tnStringToEncode,[999999999999999]))
LOCAL ARRAY laRebus[10,2]
m.laRebus[1,1]=1
m.laRebus[1,2]=[M]
m.laRebus[2,1]=2
m.laRebus[2,2]=[O]
m.laRebus[3,1]=3
m.laRebus[3,2]=[N]
m.laRebus[4,1]=4
m.laRebus[4,2]=[K]
m.laRebus[5,1]=5
m.laRebus[5,2]=[E]
m.laRebus[6,1]=6
m.laRebus[6,2]=[Y]
m.laRebus[7,1]=7
m.laRebus[7,2]=[S]
m.laRebus[8,1]=8
m.laRebus[8,2]=[H]
m.laRebus[9,1]=9
m.laRebus[9,2]=[I]
m.laRebus[10,1]=0
m.laRebus[10,2]=[P]
m.lcRetval=[]
m.lcChar=[]
FOR m.x=1 TO LEN(ALLTRIM(m.lcStringToEncode))
m.lcChar=m.laRebus[ASCAN(m.laRebus,VAL(SUBSTR(m.lcStringToEncode,x,1)),1,-1,1,8),2]
m.lcRetval=m.lcRetval+m.lcChar
NEXT
IF m.tlRandom
LOCAL m.liCounter AS Integer
PRIVATE m.pcRandomChar
m.pcRandomChar=[ ]
m.liCounter=1
*!* get a bunch of values to work with which will hopefully give at
least one alpha not in monkeyship
DO WHILE NOT ReturnRandom()
m.liCounter=m.liCounter+1
IF m.liCounter=10
EXIT
ENDIF
ENDDO
IF m.liCounter=10 && we went through loop 10 times and
could
not get a char back
m.lcRetval=[Q]+m.lcRetval
ELSE
m.lcRetval=m.pcRandomChar+m.lcRetval
ENDIF
ENDIF
RETURN m.lcRetval
ENDFUNC
FUNCTION ReturnRandom
LOCAL m.lcRandomSeed AS Character, m.lIsGood AS Boolean
*!* get a bunch of values to work with which will hopefully give at
least one alpha not in monkeyship
m.lcRandomSeed=SYS(2015)+SYS(2015)+SYS(2015)+SYS(2015)
m.lIsGood=.f.
FOR m.y=LEN(m.lcRandomSeed) TO 2 STEP -1 && don't need to
check
first char in sys(2015)
IF ISALPHA(SUBSTR(m.lcRandomSeed,y,1)) AND NOT
UPPER(SUBSTR(m.lcRandomSeed,y,1))$[MONKEYSHIP]
m.pcRandomChar=SUBSTR(m.lcRandomSeed,y,1)
m.lIsGood=.t.
EXIT
ELSE
m.lIsGood=.f.
ENDIF
NEXT
RETURN m.lIsGood
ENDFUNC
--
rk
-----Original Message-----
From: ProfoxTech <profoxtech-boun...@leafe.com> On Behalf Of Frank
Cazabon
Sent: Wednesday, October 10, 2018 3:25 PM
To: profoxt...@leafe.com
Subject: Numbering Scheme
I have a client who issues tickets in numerical sequence (it's a
pawnshop). The sequence helps them balance things back at the end of
the
day (read that as check for stealing) when checking the various
parcels
received for the tickets issued (they keep a copy of the ticket
issued to
their customer and at the end of the day sort them sequentially and
read
the numbers off the tickets to ensure they match the parcels). They
are now
not wanting the number printed on the ticket as their competitors may
be
able to get an idea from the sequential numbers how much business
they are
doing (by getting a ticket early in the morning and then one ate in
the
afternoon).
So, they have asked me to come up with a solution and I must admit
that
I am coming up blank.
Any ideas?
