Bash script to compare two lists

Shell Programming and Scripting
1

Hi,

I do little bash scripting so sorry for my ignorance.

How do I compare if the two variable not match and if they do not match run a command.

I was thinking a for loop but then I need another for loop for the 2nd list and I do not think that would work as in the real world there could be >= 100 entries in the variables.

$VAR1="me you him her"
$VAR2="you me him"

If VAR2 does not contain an entry in VAR1 create this missing entry and do xyz else do yz.

2

Given you're using bash, you could store your lists in arrays:

The following output:

zsh-4.3.11[t]% bash s
before ...
me      you
you     me
him     him
her     
after ...
me      me
you     you
him     him
her     her

... is produced by the code below:

list1=(
  me you him her
  )
list2=(
  you me him
  )

printf 'before ...\n'

paste <( printf '%s\n' "${list1[@]}" ) <( printf '%s\n' "${list2[@]}" )  
  
for i in "${!list1[@]}"; do
  [[ ${list1} != ${list2} ]] &&
    list2=${list1}
done

printf 'after ...\n'

paste <( printf '%s\n' "${list1[@]}" ) <( printf '%s\n' "${list2[@]}" )
3

POSIX shell:

VAR1='me you him her'
VAR2='you me him'
for w in $VAR1; do                                                         
  if ! echo "$VAR2" | grep -q "$w" ; then
    changed=1
    VAR2="$VAR2 $w"
  fi
done
if [ "$changed" ]; then
  echo "changed: VAR1='$VAR1' VAR2='$VAR2'"
else
  echo "no changed: VAR1='$VAR1' VAR2='$VAR2'"
fi
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