CODEWITHSUNDEEP
bash
Aditya Bhave
Aditya Bhave

Reputation: 1

Compound conditional in a bash while loop

Im modifying an existing bash script and having some trouble getting the while loop behaving correctly. This is the original code

while ! /usr/bin/executable1
do
    # executable1 returned an error. So sleep for some time try again
    sleep 2
done

I would like to change this to the following

while ! /usr/bin/executable1 && ! $(myfunc)
do
    # executable1 and myfunc both were unsuccessful. So sleep for some time
    sleep 2
done

executable1 returns 0 on success and 1 on failure. I understand that "true" in bash evaluates to 0 so thats why the original script would keep looping till executable returned success

Accordingly myfunc is coded like this

myfunc ()
{
    # Check if file exists. If exits, return 0, If not, return 1
    if [ -e someFile ]; then
        return 0
    fi

    return 1 
 }

I notice that the my new while loop does not seem to call executable1. It always calls myfunc() and then exits out of the loop immediately. What am I doing wrong?

I tried various ways of coding the while loop (with (( )), [ ], [[ ]] etc), but nothing seems to fix it

Upvotes: 0

Views: 155

Answers (1)

chepner
chepner

Reputation: 532368

You don't need $(...) to call a function, just to capture its standard output. You simply want

while ! /usr/bin/executable1 && ! myfunc
do
    sleep 2
done

Note that myfunc can also be more simply written

myfunc () {
    [ -e someFile ]
}

or even (in bash)

myfunc () [[ -e someFile ]]

Either way, it's almost not worth defining myfunc separately; just use

while ! /usr/bin/executable1 && ! [[ -e someFile ]]
do
    sleep 2
done

It might also be simpler to use an until loop:

until /usr/bin/executable1 || [[ -e someFile ]]; do
    sleep 2
done

Upvotes: 1

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