RE: Read same file -- COBOL400-L

If the CL calling PGMA, is the top pgm, then you are NOT running a COBOL 
RUN UNIT, where files once opened, stay open until closed, no matter how 
many pgm-calls you are doing.

Check for open of the file in PGMB and PGMC.  If the file is opened every 
time these pgms are called, the processing will take unneccessary long 
time.



Mvh.

Geir Kildal
TietoEnator AS
Tlf . . : (+47) 611 73805
Mob. : (+47) 90 10 13 16
eMail: geir.kildal@xxxxxxxxxxxxxxx




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15.04.2005 04:01
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COBOL Programming on the iSeries/AS400 <cobol400-l@xxxxxxxxxxxx>


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RE: [COBOL400-L] Read same file






Sorry, the information in my example is not clear. I will describe my
example more detail like this :
1. PGMA called by CL program 
   OVRDBF FILE(FILEA) TOFILE(*LIBL/FILEA) MBR(*FIRST) SHARE(*NO) 
   OPNDBF FILE(FILEA) OPTION(*ALL) TYPE(*PERM) 
   CALL   PGM(*LIBL/PGMA)

2. PGMB and PGMC used same file and read FILEA by key for different 
purpose,
like a function that return a value.
   In PGMB & PGMC not only read FILEA, but access some files and perform
something like calculation or another process.
 
Basicly, we can put procedure in PGMB and PGMC into PGMA. But if we do 
this,
then PGMA will be a big program and difficult to maintained.

Is the statement overriding file with share option equal 'NO' cause the
process to be slowly ? Because in my understanding, if share option = NO,
then for every program that open that file will be create new buffer.
For your information, we have about 200.000 record in FILEA.

Btw, Thank you very much for your advise .

rgds
Agus R


-----Original Message-----
From: cobol400-l-bounces@xxxxxxxxxxxx
[mailto:cobol400-l-bounces@xxxxxxxxxxxx]On Behalf Of
cobol400-l@xxxxxxxxxxxx
Sent: Thursday, April 14, 2005 9:37 PM
To: COBOL Programming on the iSeries/AS400
Subject: R: [COBOL400-L] Read same file


You may try to declare three times the same file in the same program, and
treat it as you like best.
E.g. 
  IDENTIFICATION DIVISION.
       PROGRAM-ID.  UNIQUE.

       SELECT FILEA-SEQ ASSIGN TO DATABASE-FILEA
           FILE STATUS IS STATUS-AS
           ACCESS MODE IS SEQUENTIAL
           ORGANIZATION IS SEQUENTIAL.
 
       SELECT FILEA-DYN ASSIGN TO DATABASE-FILEA
           FILE STATUS IS STATUS-AD
           ACCESS MODE IS DYNAMIC
           ORGANIZATION IS INDEXED
           RECORD KEY IS EXTERNALLY-DESCRIBED-KEY.

       SELECT FILEA-KEY
           ASSIGN TO DATABASE-FILEA
           FILE STATUS IS STATUS-AK
           ACCESS MODE IS DYNAMIC
           ORGANIZATION IS INDEXED
           RECORD KEY IS EXTERNALLY-DESCRIBED-KEY.

It seems that third file is treated same way than 2nd, but it problably
does'nt care.
It would be ACCESS RANDOM, may be ? 

In the program's body  you may the code:
 perform READ-FILEA-SEQ

 perform START-FILEA-DYN
 perform READ-FILEA-DYN-NXT
 perform READ-FILEA-DYN-key

and so on....



START-FILEA-DYN
                 start FILEA-DYN key not less externally described key 
invalid key
continue end-start

READ-FILEA-DYN-NXT
                 read filea-dyn next at end continue end-read

READ-FILEA-DYN-key
                 read filea-dyn key externally-described-key invalid key 
continue
end-read


Sincerely
                 Domenico Finucci
                 Sistemi informativi (SVI/SSI)
                  Tel. + 39 02-43.01.2494, cell. 348 - 59.53.279
                  Fiditalia S.p.A. via G. Silva, 34 - 20149 Milano
                 E-mail: domenico.finucci@xxxxxxxxxxxx 
> Sito internet: www.fiditalia.it
> 



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