bash how to get async command output to variable

Question

i want to know how can we get command output to variable in bash

here is my code

#!/bin/bash
opt="svcrack -v -u100 -r1-9999 -z3 10.0.0.1"
opt2="$($opt)"

echo "myout output $opt2"

output

myout output

not working inside function :(

function zwork(){

opt=$(svcrack -v -u100 -r1-9999 -z3 10.0.0.1 2>&1)

echo "myout output $opt"
}

out=$(zwork)

I don't see anything asynchronous in your script. If that's the output you get, then your command just has no output on stdout. — 4ae1e1
– 4ae1e1, Commented Nov 12, 2015 at 6:34
I said no output on stdout. Judging from your "it worked" comment below, the output is on stderr, which doesn't contradict my diagnosis. But you should see output on stderr printed nevertheless if you're running the exact script as above in an interactive shell. — 4ae1e1
– 4ae1e1, Commented Nov 12, 2015 at 6:41
As an aside, variables are for data, not code. Either define a function named opt, or use the svcrack command directly in the command substitution as in the accepted answer. See I'm trying to put a command in a variable, but the complex cases always fail!. — chepner
– chepner, Commented Nov 12, 2015 at 12:11
Came late, but I guess you had to declare your function just as zwork(){ ... } instead of function zwork(){ ... } — AngeLOL
– AngeLOL, Commented Nov 7, 2022 at 19:47

Daniel Walter · Accepted Answer · 2015-11-12 06:38:42Z

1

Please try redirecting stderr to stdout like:

#!/bin/bash
opt=$(svcrack -v -u100 -r1-9999 -z3 10.0.0.1 2>&1)

echo "myout output $opt"

Here you can read more about command substition.

edited Nov 12, 2015 at 6:38

answered Nov 12, 2015 at 6:35

Daniel Walter

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1 Comment

wayne ellis Over a year ago

sir can u please check , its not working inside function , question update

Kulvinder Singh · Accepted Answer · 2015-11-12 07:02:29Z

0

If you store a command in a variable in bash you need to use the eval keyword to execute that command. So in your case you should do something like this...

opt="svcrack -v -u100 -r1-9999 -z3 10.0.0.1"
opt2=$(eval $opt)
echo "myout output $opt2"

Correction : I was testing this in ZSH (not pure bash), so i needed to add the eval and ended up assuming that is the missing bit. However as someone corrected me in comment. That is not necessary.

Only problem in the OP script was

opt2="$($opt)"

it should have been

opt2=$($opt)

edited Nov 12, 2015 at 7:02

answered Nov 12, 2015 at 6:36

Kulvinder Singh

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1 Comment

4ae1e1 Over a year ago

Not true. Parameter expansion goes before command substitution, so the original script is valid, albeit quite stupid.

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