SORT OUTREC - Reformatting a register from dfsort parameters

[filikindri]

I've got this sort:

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//SORT    EXEC PGM=ICETOOL,                                
//        PARM='JP1"&PARA1",JP2"&PARA2",JP3"&PARA3"',                  
//        REGION=0M                                        
//TOOLMSG   DD SYSOUT=*                                    
//DFSMSG    DD SYSOUT=*                                    
//IN        DD DSN=EDI.HDRSMPL,DISP=SHR            
//OUT       DD DSN=.EDI.HEADER.R&PARA1..&PARA2,     
//         DISP=(NEW,CATLG,DELETE),                        
//         UNIT=(DISK),SPACE=(TRK,(1,1),RLSE)              
//TOOLIN DD DSN=TEPDXXX.XXX.JCLLIB(TOOLIN),DISP=SHR        
COPY FROM(IN)  TO(OUT) USING(CTL1)
//CTL1CNTL DD DSN=TEPDXXXM.XXX.JCLLIB(CTL1CNTL),DISP=SHR    
SORT FIELDS=COPY                                
OUTREC IFTHEN=(WHEN=(19,4,CH,EQ,C'FECH'),       
        OVERLAY=(19:&DATENS=(DM4),TRAN=HEX)),   
       IFTHEN=(WHEN=(23,6,CH,EQ,C'RRRRRR'),     
        OVERLAY=(23:C'ABCDEF')),                
       IFTHEN=(WHEN=(29,6,CH,EQ,C'SSSSSS'),     
        OVERLAY=(29:C'123456'))                 

(I'm using PDS members for DFSORT statements, so I paste the content after the jcl line.)

My sample file is:

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TEXTTEXTTEXT..... FECHRRRRRRSSSSSS   N..
ECEEECEEECEE220004CCCCDDDDDDEEEEEE444D23
3573357335730422006538999999222222000527

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PARA1='007302',PARA2='400'

a) I have to replace RRRRRR, with last four digits from PARA1 and two first digits from PARA2, so I would obtain '730240'. What is the correct way to do this?
b) I have to replace SSSSSS with PARA3. That´s no problem, I will just use JP3, "as is". PARA3 is 6 bytes long.
c) And I have to replace FECH with X'DDMMYYYY' , Using &DATENS=(DM4) is ok, but I.m not able to "pack" it. Can I use some value of TRAN= ?

I promise I have been looking for the answer in DFSORT manuals
Thanks a lot.

Perhaps it wuold be better two steps, using a temporal file?
COPY FROM(IN) TO(TMP) USING(CTL1)
COPY FROM(TMP) TO(OUT) USING(CTL1)

Edit: I could append PARA1+PARA2 in the first sort. In the second one , I would use BUILD to pick the correct bytes.
Packing the date, I can't see any path to do that, please, help me with this.

[filikindri]

I'm very near from solution .

Two steps sort:

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SORT FIELDS=COPY                                  
OUTREC BUILD=(1,18,&DATENS=(DM4),JP1,JP2,JP3,35,6)

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SORT FIELDS=COPY                  
OUTREC BUILD(1,18,19,8,29,6,36,12)

Now I just need to convert "19,8' to packed., so my final lenght would be 40 again instead of 44.

I remember that using 'BI' I convert x'23092010' to c'23092010', but now I need the other way, from c'' to x''. I will look for in manuals.

I think to post here, even without any response, is helping me to think

[Frank Yaeger]

You can use 19,8,TRAN=UNHEX to convert from 'ddmmyyyy' to X'ddmmyyyy'.

Of course, you don't need two passes to do what you want to do. You can just use multiple IFTHEN clauses, or INREC, OUTREC and/or OUTFIL in the same pass.

Note that X'ddmmyyyy' is NOT PACKED - it is binary. PACKED would be X'0ddmmyyyyC'.

[filikindri]

Thanks Frank, one more time...