IBM Mainframe Forum

I have a file that contains a mix of EBCDIC text and HEX.
EG. 46000102D0
I am trying to use DFSORT & OUTREC to convert the chars from position 7 to 10 back to EBCDIC.
eg. 460001 }
In essence there is an existing program that unpacks the characters to decipher them.

Note that I can get from the 460001 } value to the 46000102D0 value by using the following:
OUTREC FIELDS=(1:1,6,
7:7,54,HEX,X)

What I cant do is get it back the other way.

EBCDIC and HEX are meaningless terms the way you're using them. Maybe you mean character and binary or maybe you mean something else.

I don't know what you mean by "convert the characters from position 7 to 10 back to EBCDIC"?

C'2D0' is X'F2C4F0' ... that is a character representation and a hex representation. Both are EBCDIC.

Show your input field in hex and your expected output field in hex so I can understand the conversion you're actually trying to do.

Sorry bout confusion - everything is in essence ebcdic I am really just talking about packing and unpacking as per the following.
I want to go from:
46000102D0
to:
460001 }
0D (hex)
20 (hex)
So chars 02 (position 7 & 8) go to a single char (which may or may not be printable) but is hex x'02'
& chars D0 (position 9 & 10) go to a single char x'D0'

I haven't noted the hex for the first 6 chars (haven't go access to mainframe at moment) but they will just be normal ebcdic characters.

That isn't really packing and unpacking. It's converting HEX in character representation to equivalent binary. I can't think of an easy way to do that with DFSORT. Maybe somebody else can.

With z/OS DFSORT V1R10 PTF UK90025 or z/OS DFSORT V1R12 PTF UK90026 (Oct, 2010), you can now use DFSORT's new TRAN=UNHEX function to do this kind of thing quite easily like this:



For complete details on the new functions for DFSORT and DFSORT's ICETOOL available with the Oct, 2010 PTF, see:

http://www.ibm.com/support/docview.wss? ... g3T7000242