RE: [sqr-users] Call System

["the dragon" <ceprn@hotmail.com>]

Bruce,

When you execute a command from within an sqr it executes the command in a 
spawned shell which is probably not inheriting the environment variables of 
the calling program.  At this point I do not think you're dealing with a 
permissions issue.  Find the path on mkdir (which mkdir or whence mkdir 
should work) and place that entire command in place of your single mkdir 
command and see if that helps.

hope this helps.

peace,
clark 'the dragon' willis


PSA: Salary <> Slavery.  If you earn a salary, your employer is renting your 
services for 40 hours a week, not purchasing your soul.  Your time is the 
only real finite asset that you have, and once used it can never be 
recovered, so don't waste it by giving it away.

"Time is the coin of your life. It is the only coin you have, and only you 
can determine how it will be spent.  Be careful lest you let other people 
spend it for you."

Carl Sandburg
(1878 - 1967)

----Original Message Follows----

I have an SQR that I am running from the UNIX command line.  I have a
number off call system statements, and some of them are working.  The
following sequence if bahaving oddly:

     let $command_line  = 'mkdir -p ' || $target
      show 'command is '
      show $command_line
      call system  using  $command_line #status

The command line that displays, when copied from the log and pasted into
the UNIX command line, exeucutes as expected.  When the SQR exucutes the
command, I get the following:

mkdir:  not found

I do not believe that this is permissions because I execute the SQR from
the UNIX prompt, then execute the pasted command from the same UNIX prompt.
So the SQR execution has the same permissions and environment variables.
The value of #status is 127.  I have other call  system commands is the
same SQR that execute as expected.

Any help would be appreciated.


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