Best way to pull a matching string from an array using bash

Question

A sample array I have like below;

OVGARR=(ACCS ELB NOVOVG SCN MISOVG)

The position of the strings with *OVG can be in any index and there are more than different types of OVG strings. I want to get the value of the 1st *OVG string on the array, cut the "OVG" part of the string and assign it to a variable.

My initial solution is like below;

OV=`printf '%s\n' "${OVGARR[@]}" | grep -P 'OVG$' | head -1 | awk -F'OVG' '{print $1}'`

Though it works, is there a shorter and more efficient approach to what I would like to do?

Raman Sailopal · Accepted Answer · 2021-04-06 09:07:04Z

1

Based on your original awk solution, but stripping the need to use grep and head:

 OV="$(awk -F 'OVG' '/OVG/ { print $1;exit }' <(printf '%s\n' "${OVGARR[@]}"))"

answered Apr 6, 2021 at 9:07

Raman Sailopal

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1 Comment

Dren Over a year ago

This solution do appear to be more simple than what I have. I will try this. Thank you Raman.

choroba · Accepted Answer · 2021-04-06 09:05:46Z

0

Not shelling out seems to be much faster. How easy it is to read is subjective.

for o in "${OVGARR[@]}" ; do
    if [[ $o == *OVG ]] ; then
        OV=${o%OVG}
        break
    fi
done

answered Apr 6, 2021 at 9:05

choroba

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