Capturing a return value and output from bash function in a loop

Question

Im a little confused, I have a function that returns a exit code value and is also returning output, from running two expect scripts (I produce an exit code from the first script, and output from the second, in the function).

I am trying to figure out how use this function in a loop. I would like to loop the function until the exit code is zero, and then store the output value in a variable. Im a little stumped how to do this, I wonder if anyone can advise? My goal is:

run_expect_scripts()
{
    expect expect_script1
    exit_value="$?"
    output=$(expect expect_script2)
    echo "$output"
    return "$exit_value"
}


until [ run_expect_scripts -eq 0 -o a_counter -eq 5 ]; do
    a_counter=$(expr a_counter + 1)
    sleep 2s
done
use output of run_expect_scripts

Please how can I do this better?

user000001 · Accepted Answer · 2018-10-14 15:51:28Z

1

One way of dealing with this is to use a temp file:

#!/bin/bash

myfunc() {
   echo "output $RANDOM"
   return $((RANDOM % 5))
}

tmp=$(mktemp)
until myfunc > "$tmp" || (( count++ == 5)); do :; done
cat "$tmp"

Or if you don't want a temp file:

#!/bin/bash

myfunc() {
   echo "output $RANDOM"
   return $((RANDOM % 5))
}

ec=1
until [ $ec -eq 0 ] || (( count++ == 5)) 
do
  output=$(myfunc)
  ec=$?
done
echo "$output"

edited Oct 14, 2018 at 15:51

answered Oct 14, 2018 at 15:46

user000001

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