take value of argument into array as an index number

Question

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

mklement0 · Accepted Answer · 2015-10-29 21:18:18Z

1

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.

edited Oct 29, 2015 at 21:18

answered Oct 29, 2015 at 21:12

mklement0

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3 Comments

karabugra05 Over a year ago

I have lower array which has romen numerals from i to xxix and when I give 14 (for example ) as second argument I want to reach lower[14] not lower[2] which is given by lower[$i]

mklement0 Over a year ago

@karabugra05: I'm not sure I fully understand, but I think that's what my solution is doing - please try it.

karabugra05 Over a year ago

i am grateful to you,it happened,I achived my goal

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