CODEWITHSUNDEEP
bash
Matias Angeluk
Matias Angeluk

Reputation: 77

how do I create an array with dynamic name in bash

Example

I want to get the size of each arrangement. I have a multidimensional arrangement.

array=("SRV_1=(e1 e2 e3 e4)" "SRV_2=(e1 e2)")

for elt in "${array[@]}";do eval $elt;done

CANT_SRVS="${#array[@]}

for ((i=1;i<=$CANT_SRVS;i++));do
  CANT_E="${#SRV_$i[@]}"    <------ ERROR
  echo $CANT_E          <------- length array
done

Upvotes: 1

Views: 67

Answers (2)

Charles Duffy
Charles Duffy

Reputation: 295288

A nameref can be pointed at multiple variables; thus making srvVar refer to any of your multiple arrays below:

srv_1=(e1 e2 e3 e4)            # I don't condone the original "eval" pattern, and no part of
srv_2=(e1 e2)                  # the question hinged on it; thus, not reproducing it here.

declare -n curSrv
for curSrv in "${!srv_@}"; do  # iterates over variable names starting with "srv_"
  echo "${#curSrv[@]}"         # ...taking the length of each.
done

See this running at https://ideone.com/Js28eQ

Upvotes: 4

glenn jackman
glenn jackman

Reputation: 246744

Charles has very good advice.

A tangent on your code: in place of eval, you can use declare when you have a variable assignment saved in a variable.

value="SRV_1=(e1 e2 e3 e4)"
declare -a "$value"
varname=${value%%=*}
declare -p "$varname"

declare -a SRV_1='([0]="e1" [1]="e2" [2]="e3" [3]="e4")'

And, as Charles demonstrates, namerefs for working with the array: declare -n a=$varname

Upvotes: 0

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