Generating sequence numbers in COBOL

lakshmi_p01 · New User Joined: 28 Sep 2005 Posts: 7

I have a field which is numeric and of 3 character length. This field is getting generated automatically now. This take value from 0 to 999. So the file can have only 1000 options now. We want to have more number of options for this field so that file can take more number of records. We are converting this field to alphanumeric which is again of 3 characters. It can take from A - Z and 0 - 9 but no special characters. In mainframes since data stores in EBCDIC format A to Z comes before 0 to 9.
So the file can have AAA, AAB,.....,AA0, AA1,.....AA9, ABA,.... ZZZ,000,....999. How to generate this sequence in COBOL?

Kindly help me out.

Thanks in advance for the help.

Regards,
Lakshmi

prav_06 · Posted: Tue Jan 23, 2007 3:12 pm

Hi Lakshmi,
Its a very intresting piece of requirement but got to use lot of logic, Will tell a logic but i am not sure whether it is possible with cobol, U said that u are begining the series with AAA which is PIC X(3), if u split this variable u would be getting three individual elements, all these individual values would be having a unique ASCII value say for examle 63 for A (Not Sure??), u can increment this value to get the next alphabet i.e.. B (63+1), and once the last character reaches Z there should be a checking crtiteria which would give the next value as 0 (AAZ to AA0) you should also be knowing the ASCII value of number's to code this logic and if u get the end of series like AA9 then the program should consider the second byte from left and increment its ASCII value to make it as ABA then the series would follow till AB9 till 999, I have not tried to use the ASCII register to code COBOL program's but i know that it can be done in C language , If you could find some way to code this logic in COBOL please let us know

Cheer's,
Thamilzan.

Aji · New User Joined: 03 Feb 2006 Posts: 53 Location: Mumbai

Please check this one.

working-storage section.
01 a pic x(3) value "AAA".
01 b pic x value spaces.
01 c pic x value spaces.
01 d pic x value spaces.
01 m.
02 n occurs 3 times.
03 ws-temp pic x.
01 i pic 99 value 0.
procedure division.
p1.
move a to m.
display m.

p2.

move ws-temp(3) to b.
if b not = 9
inspect b converting
"ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789" to
"BCDEFGHIJKLMNOPQRSTUVWXYZ0123456789A"
move b to ws-temp(3)
else
move ws-temp(2) to c
inspect c converting
"ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789" to
"BCDEFGHIJKLMNOPQRSTUVWXYZ0123456789A"
move c to ws-temp(2)
move "A" to ws-temp(3).

if ws-temp(2) = 9 and ws-temp(3) = 9
display m
move ws-temp(1) to d
inspect d converting
"ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789" to
"BCDEFGHIJKLMNOPQRSTUVWXYZ0123456789A"
move d to ws-temp(1)
move "A" to ws-temp(2) ws-temp(3).
display m.
go to p2.

Aji Cherian

William Thompson · Posted: Tue Jan 23, 2007 4:01 pm

shaikmf · New User Joined: 06 May 2005 Posts: 3 Location: Hyderabad

Aji,

Good solution.

sandeep1dimri · New User Joined: 30 Oct 2006 Posts: 76

Hi
Check this pseudo code
Array 1 : ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789
ARRAY 2 : ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789
ARRAY 3 : ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789
NESTED PERFORM

perform 36 times (i.e on ARRAY1)-- COUNT-1
PERFORM 36 TIMES ON ARRAY 2---- cOUNT2
PERFORM 36 TIMES ON ARRAY 3 --- COUNT3
STRING ARRAY1(COUNT1)
ARRAY1(COUNT2)
ARRAY1(COUNT3) IN WS-SEQ
dISPAY WS-SEQ
END-PERFORM
END-PERFORM
END-PERFORM

SO WE ARE ONLY LEFT WITH COUNTING 1 TO 999
thats just a addition
Please let me know if u stuck while implementation.

Sandeep

William Thompson · Posted: Tue Jan 23, 2007 7:43 pm

William Thompson · Posted: Tue Jan 23, 2007 7:46 pm

sandeep1dimri · New User Joined: 30 Oct 2006 Posts: 76

Hi William

yeah we can do with the one array that though came to me while i left the day!!!

How would you handle a situation where you have the prior sequence number and need to generate the next one? Like given "D8K", how would you generate "D8L"?

hi i think it will also come up for ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789
ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789
ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789
and third qualifier will just proceed with K then L then ------ upto 9
so it will generate all

Please let me know if it still have logical problem
sandeep

rajesh_mbt · New User Joined: 27 Mar 2006 Posts: 97 Location: India

Aji · New User Joined: 03 Feb 2006 Posts: 53 Location: Mumbai

I have made some changes in the previous code. But it functions like the earlier one.
inspect b converting
"ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789" to
"BCDEFGHIJKLMNOPQRSTUVWXYZ0123456789A".

In the above case if the value of b is "A" after 'inspect' b becomes "B"
Similarly if b is "Z" after 'inspect' b becomes "0".

working-storage section.
01 a pic x(3) value "AAA".
01 b pic x value spaces.
01 m.
02 n occurs 3 times.
03 ws-temp pic x.
procedure division.
p1.
move a to m.
display m.
perform p2 thru p-exit.
stop run.
p2.
move ws-temp(3) to b.
if b not = 9
perform c-para
move b to ws-temp(3)
else
move ws-temp(2) to b
perform c-para
move b to ws-temp(2)
move "A" to ws-temp(3)
end-if.
if ws-temp(2) = 9 and ws-temp(3) = 9
display m
if m = 999
go to p-exit
end-if
move ws-temp(1) to b
perform c-para
move b to ws-temp(1)
move "A" to ws-temp(2) ws-temp(3)
end-if.
display m.
go to p2.
c-para.
inspect b converting
"ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789" to
"BCDEFGHIJKLMNOPQRSTUVWXYZ0123456789A".
p-exit.
exit.

Aji Cherian

William Thompson · Posted: Wed Jan 24, 2007 4:41 pm

Aji,
Your code would be much more readable if you wrapped it with "code". That will retain the indent spaces.

Aji · New User Joined: 03 Feb 2006 Posts: 53 Location: Mumbai

Thanks for the suggestion William.

Aji Cherian

William Thompson · Posted: Wed Jan 24, 2007 4:51 pm