How to copy an array to a new array with dynamic name?

Question

I have a complex data structure in Bash like this:

Test1_Name="test 1"
Test1_Files=(
  file01.txt
  file02.txt
)
Test2_Name="test 2"
Test2_Files=(
  file11.txt
  file12.txt
)
TestNames="Test1 Test2"

In my improved script, I would like read the files from disk. Each test resides in a directory.

So I have a Bash snippet reading directories and reading all the file names. The result is present in an array: $Files[*].

How can I copy that array to an array prefixed with the test's name. Let's assume $TestName holds the test's name.

I know how to create a dynamically named array:

for $Name in $TestNames; do
  declare -a "${TestName}_Files"
done

Will create e.g. Test1_Files and Test2_Files. The variables are of type array, because I used -a in declare.

But how can I copy $Files[*] to "${TestName}_Files" in such a loop?

I tried this:

declare -a "${TestName}_Files"=${Files[*]}

But it gives an error that =file01.txt is not a valid identifier.

ilkkachu · Accepted Answer · 2018-12-31 13:34:35Z

2

You'll want to use newarray=( "${oldarray[@]}" ) to keep the array elements intact. ${oldarray[*]} will involve word splitting, which will break at least with elements containing white space.

However, the obvious declare -a "$name"=("${oldarray[@]}") doesn't work, with the parenthesis quoted or not. One way to do it seems to be to use a name ref. This would copy the contents of old to new, where the name of new can be given dynamically:

#!/bin/bash
old=("abc" "white space")
name=new

declare -a "$name"        # declare the new array, make 'ref' point to it
declare -n ref="$name"    

ref=( "${old[@]}" )       # copy

#unset -n ref             # unset the nameref, if required
declare -p "$name"        # verify results

edited Dec 31, 2018 at 13:34

answered Dec 31, 2018 at 12:16

ilkkachu

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2 Comments

Paebbels Over a year ago

What is the purpose of declare -p "$name"?

ilkkachu Over a year ago

@Paebbels, just to print out the new copy to verify the result is correct

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