Bash shell operator {1..n} not working when using variables

Asked Feb 01 '15 at 16:38

Active Feb 01 '15 at 16:40

Viewed 23 times

I am working on a progress bar, so I need to print several consecutive # characters. I am using Bash shell operation.

This is the case example:

$ a=20
$ printf "#%.0s" {1..a}
#$ printf "#%.0s" {1..$a}
#$ printf "#%.0s" {1.."$a"}
#$ printf "#%.0s" {1..20}
####################

Note that it is only generated one only #, except when using the direct 20 value in the command line. Invoking the a variable seems not to work.

I need to do this programmatically, because 20 could be any other number.

What am I missing? Something obvious/stupid?
Thanks you.

edited Feb 01 '15 at 16:40

Tom Fenech

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asked Feb 01 '15 at 16:38

Sopalajo de Arrierez

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Related: http://unix.stackexchange.com/questions/7738/how-can-i-use-var-in-a-shell-brace-expansion-of-a-sequence – Tom Fenech Feb 01 '15 at 16:43
Try ((a=20)). I would try it out first before suggesting it, but I am not near a computer. – Danny Daglas Feb 01 '15 at 18:18
1

printf "#%.0s" $(seq 1 $a) – fpmurphy Feb 01 '15 at 18:26
It requires `seq`, but working all right, @fpmurphy1 . Thanks. – Sopalajo de Arrierez Feb 01 '15 at 23:11

Bash shell operator {1..n} not working when using variables

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