I have problem with getting value of parameter when I have variable with some value. What I mean is:
Example:
./script 12 13 14 15 16
value=5
echo $value #now I see 5
$"$value" #now I want to get 16 but I don't know how to do it?
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4 Answers
Use indirection:
echo "${!value}"
Quotes aren't necessary for the value 16, but should be used in case the variable contains special characters.
Comments
try this as well:
value=5 #
echo "$value" # 5
echo ${@:$value:1} # give you 1 arg starting from $value in the arg list
Comments
you need to dereference that variable
value=5
echo "$value" # 5
echo "${!value}" # will give you $5 or in your example 16
Comments
bashisms are inherently non-portable. If you rely on ${!...} to evaluate the expression, your script will only run in bash. This may not be an issue, but it is an issue if the author of the script is blissfully unaware of the lack of portability. This sort of thing is trivial to do without relying of bashisms. If you want to evaluate a string, use eval:
eval echo \$$value
