[Solved]converting Integer date extracted from DB2 to 9(8) format

pranabbiswas · New User Joined: 03 Jun 2007 Posts: 3 Location: Irvine,USA

- I extracted DB2 date defined as INTEGER in a sequential file.
- Found that the the date is stored in 4 bytes as hexadecimal format. When I used calculator to convert it from hex to decimal it displays as 9(8) date e.g 20010718.
- If I want to move this hex date to a field defined in cobol as 9(8) or X(8) to display the date as the above format as converted by the calculator. How can I do that in a COBOL program.
I would appreciate any help in this regard
Thanks, Pranab

bansal · New User Joined: 03 Jan 2007 Posts: 27 Location: Hyderabad

Hi,

First of all, if u use copybook created by DCLGEN, u dont need to worry about it. Bcoz it will take the standard settings. And the standard format of DATE in COBOL is 10 bytes but not 8 bytes.

If u want it in 8 bytes, u can extract the Db2 date in 10 bytes and from there u can extract the 8 bytes to a variable.

Hope this helps

dick scherrer · Posted: Fri Jun 08, 2007 12:27 am

Hello,

Try to MOVE it to a field defined as PIC 9(8) and display that result field.

Please post your variables and the move/display if it does not work the way you want.

pranabbiswas · New User Joined: 03 Jun 2007 Posts: 3 Location: Irvine,USA

Thanks for your help.
I have defined the variable as Pic 9(8) Binary
Then moved it to the target field with Pic 9(80).
That resolved my problem.
It displayed as ccyymmdd format.
Thanks,
Pranab

dick scherrer · Posted: Fri Jun 08, 2007 1:18 pm

You're welcome

And might this

pranabbiswas · New User Joined: 03 Jun 2007 Posts: 3 Location: Irvine,USA

Yes, you are right. It was a typo.
Was moved to 9(8).

Thanks a lot!

dick scherrer · Posted: Sat Jun 09, 2007 10:14 am

You're welcome