CODEWITHSUNDEEP
bash
karabugra05
karabugra05

Reputation: 55

take value of argument into array as an index number

argument_number=$#
for ((i = 2; i <= $argument_number; i++))
do
echo ${lower[$i]]}
done

I want to take argument value into array as index number for example if my second argumnet is 15 want to achive lower[15] but it gives lower[2] how can I do ,is there any suggestion,help

Upvotes: 0

Views: 44

Answers (1)

mklement0
mklement0

Reputation: 437373

If I understand you correctly, the indices into array ${lower[@]} are given by command-line arguments; in this case, you can simply use:

shift # skip the 1st argument
for index; do # this loops over all command-line arguments; same as: for index in "$@"; do
  echo "${lower[index]}"
done

Note:

  • I've double-quoted ${lower[index]} so as to protect the value from unwanted interpretation by the shell - unless you specifically want the shell to perform word-splitting and globbing on a variable value, you should double-quote all your variable references.

  • Array subscripts in Bash are evaluated in arithmetic context, which is why variable index can be referenced without the usual $ prefix.

Upvotes: 1

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