How to check a program run successful in bash script?

Question

I have to run some programs in bash script.Sometimes program may not give result.

#!/bin/sh
for k in $( seq 1 20)
do
echo -n ${k}' ' >> 2.txt
./s${k}>>2.txt & sleep 5
done

My ideal result like this:

1 result001
2 result002
3 
4 result004

But after run my code,the result is

1 result001
2 result002
3 4 result004

So my problem is how to check a program run successful in bash script?Thank you!

Answer 1

This should work for you.

 #!/bin/bash


    for k in {1..20}
    do
       result=$(./s${k} & sleep 5)
       printf "%b %b\n" $k $result >> 2.txt
    done

The following output was given when testing this script:

1 result1
2 result2
3 
4 result4

Answer 2

Try to use wait

http://en.wikipedia.org/wiki/Wait_%28command%29

wait [n]

where n is the pid or job ID of a currently executing background process (job). If n is not given, the command waits until all jobs known to the invoking shell have terminated.

wait normally returns the exit status of the last job which terminated. It may also return 127 in the event that n specifies a non-existent job or zero if there were no jobs to wait for.

Because wait needs to be aware of the job table of the current shell execution environment, it is usually implemented as a shell builtin.

Just add wait after this line

./s${k}>>2.txt & sleep 5
wait

In case the process can deadlock - another option is timeout

timeout -s 9 5 ./s${k}>>2.txt # Waits 5 seconds and then kill the process

Answer 3

You could call the jobs sequentially and use timeout to limit execution time:

#!/bin/sh
for k in $( seq 1 20)
do
  echo -n ${k}' ' >> 2.txt
  timeout -k 8 5 ./s${k}>>2.txt
done

This will terminate each job with the TERM signal (allowing cleanup via signal handlers to run) after 5 seconds and, if unsuccessful, with the KILL signal after 8 seconds.

However from your shell's perspective, no parallel processes are involved and you don't need to solve any concurrency problems.