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Suppose there is a table like the following...I just wanted to know if there is any command using which we can get the record/name of the person who joined before 2005..


Sl Name des y.o.joining...

Actually the main script which i posted yesterday is saimptlogi_batch.sh and i am submitting this script in a job scheduler call "AppWorx" ....What this script does is it consume some file called...

Please anyone can help me out in this one....I need this very urgently....

#!/usr/bin/ksh
#!/usr/bin/perl
set -x
environment=rms
export environment
#. $AW_HOME/RETEK/exec/RETEK_ENVAR
ls -ltr $MMPOS/RTLOG* | tr -s " " | cut -d " " -f9...

What this "do" will do ???

#!/usr/bin/ksh
#!/usr/bin/perl
environment=rms
export environment
. $AW_HOME/RETEK/exec/RETEK_ENVAR
ls -ltr $MMPOS/RTLOG* | tr -s " " | cut -d " " -f9 >$MMHOME/oracle/proc/bin/rtlogfiles.txt...

Thanks...I will test it and inform you abt the result....

Sir....It is giving the following error.....

savouch_75002_20090210_20090226160640.out/appdb/product/agent/tmp/AAAjzaWJcsyntax error at -e line 1, near "; exit"
Execution of -e aborted due to...

So instead of sleep 140 in the above mentioned script.....can I put it like this..will it work?:

cd $MMHOME/log
while
do
more $(ls -ltr|tail -1|cut -c 55-70)|grep "saimptlogi"|tail -2f|perl...

BRO...wait $! is probably not the right solution and it's not working also for this particular script

What I need is instead of sleep I want few lines of code which will check for the PID in the...

i am curious to know how wait $! will identify the PID ....because at the same time many other jobs will also be running....

Hey but i think in wait we have to mention the PID but here each and every job will be having different PID ..but the PPID will be same for all that is supp.6415......Since I am not an Unix expert...

Bro...its not working....
I dont want the sleep option atall...
Currently because of this sleep140 even if the first job gets complete in one minute it is unnecessarily waiting for the next job.......

Thanks pludi (https://www.unix.com/members/302054671.html) and Praveen ....

Both the scripts are working.....

In pludi (https://www.unix.com/members/302054671.html)'s solution i added just one...

Hey Guys....
Just need some help as I am not proficient in Unix shell script...
Doubt:
---------------
Suppose there will be some of the following files inside a directory called OUT ...

Path:...

It's not working sir..Any other option...

Suppose I have a directory called jeet and inside that directory so many files will be there....
Example:
/abc/xyz/jeet
$ ls -ltr
total 0
-rw-r--r-- 1 oracle dba 0 Jan 13 11:36...

I am new to Shell Scripting and I need help to write the following script in a different format...

This is the current script:

#!/usr/bin/ksh
environment=rms
export environment
....