Shell Programming and Scripting

Hi,

I have a main script(main.ksh) within which I have called another script(sub.ksh). The sub.ksh script is made to run in the background using '&'. The main.ksh script logs the information in a logfile main_ddmmyy and the sub.ksh script also logs the information in the log file sub_ddmmyy. This sub.ksh script will take long time to execute.

My problem arises here. Since the sub.ksh script is executed at the background,I started another process(new.ksh) in the foreground. The new.ksh also logs the information in main_ddmmyy file . The new.ksh script is executed properly and logs the information in its log file, but the sub.ksh script is not logging the information in its logfile when I have process both in foreground and background. If I don't start the process(new.ksh),the information from the background script(sub.ksh) is logged properly.

Hope I have made you clear.

I am not able to figure out where the problem is. Please help me on it.

Thanks in advance
Chella

It sounds like the two functions are logging to the same logfile?
If so, make sure both are using >> rather than > so the entries get interleaved.

Alternativly, give them different logs to write to.

Thanks for the reply.

I don't have any problem in logging into same file. But the background process fails to log the information in its log file when I have a foreground process running which logs the information in different file. I want to make clear that the foreground and background process are not logging the information in the same file.
Main.ksh - Main_ddmmyy(Foreground)
Sub.ksh - Sub_ddmmyy(background process)
New.ksh - Main_ddmmyy(Foreground)
The problem is with logging the sub_ddmmyy file.

Please suggest me on it.

Thanks in advance.
Chella

can we see some code as to how you are doing your logging??

Note that this will not work:

Code:

sub.ksh & >/tmp/logfile 2>&1

but this will:

Code:

sub.ksh >/tmp/logfile 2>&1 &

Redirection can get tricky, which is why I add this code to any of my scripts that run in the background or unattended:

Code:

exec >/tmp/${0##*/}.log 2>&1

After this line, all output (stdout and stderr) will go to a file in /tmp using the program basename with a ".log" extension. Don't try this from the command line!

UPDATE: I actually use a base set of code for background jobs:

Code:

# Send all output to a logfile and supress input
typeset LOG="/tmp/${0##*/}.out"
mv $LOG ${LOG}.old >/dev/null 2>&1
[[ -t 1 ]] && echo "Writing to logfile '$LOG'."
exec > $LOG 2>&1
exec < /dev/null 2<&1

Note that the shell expression "${0##*/}" is the same as "basename $0". Also, the "[[ -t ]]" should return TRUE when the process is attached to a terminal. In this case, it just prints a message about the logfile (instead of having the user wonder if the script is even running).

I also close stdin since an unattended script should never need input. For instance, if you have "grep $FILENAME" or "cat $FILENAME" and a bug has left $FILENAME unset, then the script would normally hang waiting for input from stdin. If that input is already /dev/null then the script can continue.

Thanks a lot for the reply.