How to get printf using {1..$VAR} to work properly

Question

When I do this, I get what I expect:

$ printf "https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=%d\n" {1..10}
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=1
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=2
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=3
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=4
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=5
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=6
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=7
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=8
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=9
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=10

But when I do this, it seems to treat it as 1 arg:

cbongiorno at wa-christianb-mbp in ~/dev/mystuff/bashful on master [!?]
$ count=10

cbongiorno at wa-christianb-mbp in ~/dev/mystuff/bashful on master [!?]
$ printf  "https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=%d\n" {1..${count}}
bash: printf: {1..10}: invalid number
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=0

cbongiorno at wa-christianb-mbp in ~/dev/mystuff/bashful on master [!?]
$ printf  "https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=%d\n" "{1..${count}}"
bash: printf: {1..10}: invalid number
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=0

cbongiorno at wa-christianb-mbp in ~/dev/mystuff/bashful on master [!?]
$ printf  "https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=%d\n" {1.."${count}"}
bash: printf: {1..10}: invalid number
https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=0

How to I get bash to expand this arg like it does when I hard-code the number?

score 1 · Answer 1 · answered Dec 12 '19 at 12:58

1

You can rewrite your line on this way:

printf "https://api.bitbucket.org/2.0/repositories/twengg?pagelen=0&page=%d\n" $(seq 1 $count)

Command seq is often used for generation of consecutive (based on rules) numbers

answered Dec 12 '19 at 12:58

Romeo Ninov

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I actually have something similar already and others are saying seq is bad. Was kinda hoping to use something native to bash – Christian Bongiorno Dec 12 '19 at 13:29
I do not know why `seq` is described as bad. This is standard UNIX/Linux command which follow the philosophy of UNIX - one program, one task :) – Romeo Ninov Dec 12 '19 at 13:32
Well, what you have is slightly better than what I have. I will give it some time to see if someone can explain why this doesn't work (which it should) – Christian Bongiorno Dec 12 '19 at 13:41
@ChristianBongiorno, probably having variable disrupt the construction of curly brackets – Romeo Ninov Dec 12 '19 at 13:49

How to get printf using {1..$VAR} to work properly

1 Answers1