I have an array created and I want to look for a keyword and then if found, display the element that is n elements behind it. Is that possible to do with bash and if so, could someone provide an example? Example of operation:
element 1=red
element 2=blue
element 3=green
Parse through the array and if you see "green", display the element that is 2 behind it. In this case, it would return "red".
#!/usr/bin/env bash
val="blue"
array=("red" "green" "blue")
for (( i = 0; i < ${#array[@]}; i++ )); do
if [[ $val == ${array[i]} ]] && (( i - 2 >= 0 )); then
echo "${array[i - 2]}"
fi
done
Outputs:
red
As @jordanm points out in the comments, you will need to worry about what might happen if an array index hasn't been set.
[] (I don't recall the proper term), is already numerical context. "${array[i + 2]}" will work. I upvoted because this is a good enough example to get the OP started, but there are other things to worry about such as arrays being sparse and what to do if i + 2 or i - 2 expands to null.
i - 2 based on the OP's use of the word "behind". Also, negative subscripts will cause an error in Bash versions before 4.2.#!/bin/sh
val="blue"
offset=2 #offset index when echoing
array=("red" "green" "blue" "purple" "blue")
for i in $(seq $offset $[${#array[*]}-1])
do
if [ "${array[$i]}" == "$val" ]
then
echo "${array[$[$i-$offset]]}"
fi
done
All simple artihmetics can be done inside $[...]; don't forget the $ before the name of the variable. It is very useful when dealing with arrays in bash. note the offset at the beginning of the for-loop, avoiding useless tests of negative indexes.
$(seq 0 $[${#array[*]}-1-$offset]). No modification is needed elsewhere. In order to make it working with positive and negative values, I suggest you to compute before the for loop the bounds of the seq instruction, depending on the sign of the offset, and start your for loop with for i in $(seq $begin $end)