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Message-ID:	<F76C9B2DA2FC4C4CA0A18E288BBCBCF7082177E9@nihexchange24.nih.gov>
From:	"Buchbinder, Barry (NIH/NIAID)" <BBuchbinder AT niaid DOT nih DOT gov>
To:	"'Thomas L Roche'" <tlroche AT us DOT ibm DOT com>, cygwin AT cygwin DOT com
Subject:	RE: [OT?] make bash script wait for called program to finish?
Date:	Sun, 18 Apr 2004 18:56:47 -0400
MIME-Version:	1.0
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How about the following?
$ cmd /c start /wait program_being_waited_for

-----Original Message-----
From: Thomas L Roche
Sent: Sunday, April 18, 2004 6:20 PM
To: cygwin AT cygwin DOT com
Subject: [OT?] make bash script wait for called program to finish?

How to make a bash script wait for a program that it calls to finish? Why I
ask:

Don't ask me why this process is structured this way, but, in order to
install some code that I need in order to test my code, I hafta

a wget an (executable) extractor and some (non-executable) binaries

b run the extractor on the binaries. This runs some InstallShield UI
  that no one I know seems to know how to script. This creates an
  installer and a set of disk images.

c run the installer on the images. Fortunately this part I know how to
  script.

What I'd really like to do is script the whole thing, because each step
takes about 30 min, but step b eludes me. So what I'd like my script to do
is

0 wget

1 launch the extractor UI, which I will briefly twiddle

2 wait for the extractor to finish

3 check that the installer and images were created

4 run the installer on the images

I know how to script everything but 2: how can I do that?

Apologies if this is a FAQ: if so, please pass a pointer. If possible please
post to me as well as the list, and TIA.

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