Variables in bash seq replacement ({1..10})
BashSeqBash Problem Overview
Is it possible to do something like this:
start=1
end=10
echo {$start..$end}
# Ouput: {1..10}
# Expected: 1 2 3 ... 10 (echo {1..10})
Bash Solutions
Solution 1 - Bash
In bash, brace expansion happens before variable expansion, so this is not directly possible.
Instead, use an arithmetic for loop:
start=1
end=10
for ((i=start; i<=end; i++))
do
echo "i: $i"
done
###OUTPUT
i: 1
i: 2
i: 3
i: 4
i: 5
i: 6
i: 7
i: 8
i: 9
i: 10
Solution 2 - Bash
You should consider using seq(1). You can also use eval:
eval echo {$start..$end}
And here is seq
seq -s' ' $start $end
Solution 3 - Bash
You have to use eval:
eval echo {$start..$end}
Solution 4 - Bash
If you don't have seq, you might want to stick with a plain for loop
for (( i=start; i<=end; i++ )); do printf "%d " $i; done; echo ""
Solution 5 - Bash
I normally just do this:
echo `seq $start $end`
Solution 6 - Bash
Are you positive it has be BASH? ZSH handles this the way you want. This won't work in BASH because brace expansion happens before any other expansion type, such as variable expansion. So you will need to use an alternative method.
Any particular reason you need to combine brace and variable expansion? Perhaps a different approach to your problem will obviate the need for this.
Solution 7 - Bash
use -s ex: seq -s ' ' 1 10 output: 1 2 3 4 5 6 7 8 9 10