Re: Displaying a message using QmhSndPm...

[Robert Rogerson <rogersonra@xxxxxxxxx>]

Thanks John,

I did look at your example and changed my program accordingly.
I changed the Call Stack Entry = 'MYPROGRAM' and Call Stack Counter = 0.
When I ran the program no message was displayed.  When I ran the program in
debug it definately sent the message but nothing appeared on the screen.
This is the approach I'm trying to achieve (not having to specify the Call
Stack Counter for each procedure I'm in) but I'm still unable to accomplish
it.

Do you have any insight?

Thanks,

Rob

On Thu, Mar 11, 2010 at 2:13 PM, John McKay <jmckay@xxxxxxxxxxxxxxxx> wrote:

> At www.rpglanguage.com/message_subfile I have an example of what I think
> you
> are looking for.  All the parameters, except for message id and message
> queue, are defaulted and the QMHSNDPM is stored in a procedure z_sndmsg.
>
> You can contact me directly for more clarification.
>
>
> Regards,
> John McKay   mba
>
> ----- Original Message -----
> From: "Robert Rogerson" <rogersonra@xxxxxxxxx>
> To: <RPG400-L@xxxxxxxxxxxx>
> Sent: Thursday, March 11, 2010 6:07 PM
> Subject: Displaying a message using QmhSndPm...
>
>
> I don’t quite understand the Call Stack Entry and Call Stack Counter
> combination.
> What I have currently, and working, is as follows:
>
> Mainline
>  QmhSndPm( ‘XXX9999’ : ‘QCPFMSG   *LIBL’ : sndData : %Len(sndData) :
> ‘*INFO’ : ‘*’ : 1 : sndMsgKey : errorCode);
>  Process1();
>
> In Process1 I have
>  QmhSndPm( ‘XXX9999’ : ‘QCPFMSG   *LIBL’ : sndData : %Len(sndData) :
> ‘*INFO’ : ‘*’ : 2 : sndMsgKey : errorCode);
>  Process2();
>
> In Process2 I have
>  QmhSndPm( ‘XXX9999’ : ‘QCPFMSG   *LIBL’ : sndData : %Len(sndData) :
> ‘*INFO’ : ‘*’ : 3 : sndMsgKey : errorCode);
>
> So as it stands all three messages are sent to the display as expected.
> I would like to create a procedure called SendMessage with one parameter
> messageId (Char 7).
>
> Is this possible and if so what should be the values of the Call Stack
> Entry
> and Call Stack Counter combination?
>
>  Thanks,
>
>  Rob
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