how do I create an array with dynamic name in bash

Question

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

Not an exact duplicate, but you can use the technique from the answer to [this question](https://stackoverflow.com/q/53733286/3266847). — Benjamin W., Dec 12 '18 at 15:51
Aside: `eval $elt` introduces bugs you don't have with `eval "$elt"` (which still is error-prone, but not quite as much so). — Charles Duffy, Dec 12 '18 at 15:56
...to showcase a specific issue with `eval $elt` vs `eval "$elt"`, btw, try running `set -x; elt="array=( ' * ' )"; eval $elt; declare -p elt` in a non-empty directory. — Charles Duffy, Dec 12 '18 at 16:13

Charles Duffy · Accepted Answer · 2018-12-12T17:07:00.850

4

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

edited Dec 12 '18 at 17:07

answered Dec 12 '18 at 15:59

Charles Duffy

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Love the nameref control variable ;) – Benjamin W. Dec 12 '18 at 16:10
1

Yup, you brought me around on that one :) – Charles Duffy Dec 12 '18 at 16:11
thank you I was just looking for something like that and it's going to be useful for the project I'm doing I was already going crazy with trying – Matias Angeluk Dec 12 '18 at 16:14
Hmmm, using `declare -n curSrv`, `echo "${#srvVar[@]}"` isn't going to do much... (typo -- welcome to the human race `:)` – David C. Rankin Dec 12 '18 at 16:55

glenn jackman · Answer 2 · 2018-12-12T16:30:11.647

0

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

edited Dec 12 '18 at 16:30

answered Dec 12 '18 at 16:23

glenn jackman

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how do I create an array with dynamic name in bash

2 Answers2