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Executes scripts parallelly based on their success

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# 8  
Old 02-10-2014
Hi Cero ,
Thanks for reply ,
Can you plz help me the command you have suggested
Code:
wait
[ -f x.flagfile -a -f y.flagfile ] && Z.sh

this means this success condition would have to check with " if " loop ?

I have tried with below logic but failed if X and Y scripts are not finishing at same time

Code:
X.sh &
Y.sh &
wait (10)
s1_addr=`sed -n "1{p;q;}" $temp/x_success.tmp |cut -d ":" -f1`
s1_cont=`sed -n "1{p;q;}" $temp/y_success.tmp |cut -d ":" -f1`
if [ $s1_addr == "success" && $s1_cont == "success"  ];then
Z.sh

# 9  
Old 02-10-2014
Actually I like Don Craguns solution better than the one I suggested because you do not need flagfiles.
The wait command without an argument waits for all background jobs you submitted. When you give an argument it is interpreted as the process id to wait for.
I'll try to explain Don Craguns code:
Code:
# start X.sh in background
bash X.sh &
# store the process id of X.sh in the variable Xpid
Xpid=$!
# start Y.sh in foreground. If Y.sh finishes sucessfully wait for X.sh to finish.
# wait $Xpid returns the return code of the process that was waited for (X.sh)
# if this was sucessfull start Z.sh in foreground. 
bash Y.sh && wait $Xpid && bash Z.sh
# the syntax A && B && C says: start A, then if it returned success start B, then if B returned success start C

These 2 Users Gave Thanks to cero For This Post:
chandini Don Cragun
# 10  
Old 02-17-2014
Thanks Don ,
Mentioned logic is work well ..!!!
Thanks for your suggestion .
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