assign variable inside if condition in bash 4?

Question

is it possible to assign variable inside if conditional in bash 4? ie. in the function below I want to assign output of executing cmd to output and check whether it is an empty string - both inside test conditional. The function should output

"command returned: bar"

myfunc() {

local cmd="echo bar"
local output=

while [[ -z output=`$cmd` ]];
do
    #cmd is failing so far, wait and try again
    sleep 5
done

# great success
echo "command returned: $output"
}

why the above?

i prefer to run scripts with 'set -e' - which will cause script to terminate on first non-0 return/exit code that's not in an if/loop conditional.

with that in mind, imagine cmd is an unstable command that may exit with > 1 from time to time, and I want to keep calling it until it succeeds and i get some output.

Did you try it? Why do you want to do it in the first place? — Carl Norum
– Carl Norum, Commented Feb 3, 2012 at 21:24
I just rewrote it to make it clear that that is not an assignment.. — Karoly Horvath
– Karoly Horvath, Commented Feb 3, 2012 at 21:29
@yi_H sorry, don't follow you - i do want to assign value to output and check whether it is an empty string - inside test conditional — Nikita
– Nikita, Commented Feb 3, 2012 at 21:34

Dennis Williamson · Accepted Answer · 2012-02-03 22:40:49Z

3

You can try something like this:

myfunc() {

    local cmd="echo bar"
    local output=

    while ! output=$($cmd) || [[ -z output ]];
    do
        #cmd is failing so far, wait and try again
        sleep 5
    done

    # great success
    echo "command returned: $output"
}

Note that it is strongly recommended to avoid the use of set -e.

answered Feb 3, 2012 at 22:40

Dennis Williamson

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1 Comment

Nikita Over a year ago

thanks for the answer - works for me. Too bad about 'set -e' being unreliable - IMO achieving the goal behind set -e would be highly useful

vmpstr · Accepted Answer · 2012-02-03 21:40:02Z

0

I don't think you would be able to do it in your conditional

As yi_H pointed out, the if is equivalent to

if [[ ! -z output=bar ]];

which in turn is basically

if [[ ! -z "output=bar" ]];

So, all you are checking is if the string "output=bar" is empty or not...

So, output=bar could actually be anything like !@#!@%=== and it would still do the same thing (that is, the expression isn't evaluated). You might have to assign the variable in a subshell somehow, but I'm not sure that would work.

answered Feb 3, 2012 at 21:40

vmpstr

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Comments

Karoly Horvath · Accepted Answer · 2012-02-03 21:48:06Z

0

Since assignment won't work there, you need some workaroudn.

You could temporary do a set +e...

answered Feb 3, 2012 at 21:48

Karoly Horvath

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Comments

amit_g · Accepted Answer · 2012-02-07 17:17:12Z

-1

You could use this way ...

$cmd
exit_status=$?

while [[ $exit_status -gt 0 ]];
do
    #cmd is failing so far, wait and try again
    sleep 5

    $cmd
    exit_status=$?
done

EDIT: This won't work with 'set -e' or other way around, don't use 'set -e' to begin with.

edited Feb 7, 2012 at 17:17

answered Feb 3, 2012 at 22:26

amit_g

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2 Comments

Nikita Over a year ago

if 'set -e' is used (see question), and $cmd returns != 0, bash will exit after the very first line

amit_g Over a year ago

Agreed, and that is why this answer did not include it. 'set -e' causes confusion and forces to write scripts in not natural way. IMO, it is more of a trouble than it is worth. This small piece of code is the very example why it should be avoided.

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