Bash command check if string match in command output-CodePudding

I have following command output

$ /opt/CrowdStrike/falconctl -g --aid | grep 'aid='
aid="fdwe234wfgrgf34tfsf23rwefwef3".

I want to check if there is any string after aid= (inside ""). If there is any string, command return code should be 0 and if no value return code must be !=0.

Can someone please help to extend this command to get required output?

Idea is to make sure my bash script to fail if aid= doesn't has any value.

CodePudding user response：

You can use regex to check whether one or more characters exist inside the double quotes. And, you can use regex capture group to extract that value:

if [[ $(/opt/CrowdStrike/falconctl -g --aid | grep 'aid=') =~ ^aid=\"(. )\"$ ]]; then
  aid=${BASH_REMATCH[0]}
  echo "aid is $aid"
else
  echo "aid not found"
fi

Note that the regex I use is . which means 1 or more characters, since you require the string to be non-empty. This is in contrast of the usual .* regex which would have be 0 or more characters.

CodePudding user response：

I don't have falconctl on my system so to mimic its output I'll use a couple files:

$ head falcon*out
==> falcon.1.out <==
some stuff
aid="fdwe234wfgrgf34tfsf23rwefwef3".
some more stuff

==> falcon.2.out <==
some stuff
aid=""
some more stuff

One grep idea:

grep -Eq '^aid="[^"] "' <filename>

Where:

-E - enable extended regex support
-q - run in silent/quiet mode (suppress all output)
the return code can be captured from $?

Taking for a test drive:

for fname in falcon*out
do
    printf "\n############# %s\n" "$fname"
    cat "$fname" | grep -Eq '^aid="[^"] "' "$fname"
    echo "return code: $?"
done

This generates:

############# falcon.1.out
return code: 0

############# falcon.2.out
return code: 1