BASH: loop-reading a list creating a dynamic array

Question

I have a list formatted as below. If I replace the list by, say, "aaa bbb ccc", everything works fine. Anyone an idea?

list=$(echo -e "1.1 1.2 mgmt")

n=0
for i in $list; do
   let "array1_$i[$n]=$(date "+%N")"
   ((++n))
done

n=0
for i in $list; do
   var=array1_$i[$n]
   echo ${!var}
   ((++n))
done

thanks in advance for any ideas...

choroba · Accepted Answer · 2013-05-14 09:20:15Z

1

You can change the shebang line to

#!/bin/bash -xv

to see what bash tries to run. In fact, you are trying to create a variable named array1_1.1, which is not a valid variable name: dots are not allowed.

answered May 14, 2013 at 9:20

choroba

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1 Comment

kampu Over a year ago

I am amazed at the usefulness of that information. I can see what bash tries to run? UPVOTE. (On investigation, -xv also works for dash (my favored shell for scripts that have to run really fast), zsh, ksh and possibly all other major sh-variants.)

mata · Accepted Answer · 2013-05-14 09:23:38Z

1

. is not a valid character in an identifier.

man bash:

       name   A word consisting only of alphanumeric characters and underscores, and
              beginning with an alphabetic character or an underscore.  Also
              referred to as an identifier

answered May 14, 2013 at 9:23

mata

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