BASH variable values as new defined variables

Question

I don't know exactly how to ask this in English, but I want to have the value of a variable as a new variable... The script also has a loop with increasing numbers, and in the end I want to have the variables VAR1, VAR2 etc.

I'm trying this:

COUNT=$(echo 1)
DEFINE=$(echo VAR$COUNT)
$DEFINE=$(echo gotcha!)

When I try this way, I have this error message:

~/script.sh: line n: VAR1=gotcha!: command not found

I played a bit around with brackets and quotation marks, but it didn't work... any solutions?

The question is tagged with bash. Why aren't you instead using bash arrays? — John1024
– John1024, Commented Oct 31, 2014 at 18:04
Why are you using echo in a command substitution? this is really silly. Use this instead: count=1, define=var$count. Now for the last one, you may use printf -v "$define" '%s' 'gotcha!'. — gniourf_gniourf
– gniourf_gniourf, Commented Oct 31, 2014 at 18:06

choroba · Accepted Answer · 2014-10-31 18:07:00Z

2

The problem is that bash expects a command as a result of expansions, not an assignment. VAR1=gotcha! is not a command, hence the error.

It would be better to use an array:

COUNT=$(echo 1)
VAR[COUNT]='gotcha!'
echo ${VAR[COUNT]}

I guess $(echo 1) stands for a more complex command, otherwise you can just use COUNT=1.

answered Oct 31, 2014 at 18:07

choroba

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Comments

chepner · Accepted Answer · 2014-10-31 18:09:05Z

1

You can use declare to create such a "dynamic" variable, but using an array is probably a better choice.

COUNT=1
DEFINE="VAR$COUNT"
declare "$DEFINE=gotcha"

answered Oct 31, 2014 at 18:09

chepner

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1 Comment

Timon Tigerli Over a year ago

Thank you. This solved my SPECIFIC problem completely. Also thanks to all other contributors. As a neophyte, i'm surely gonna fall in love with StackOverflow ^^.

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