Variables in bash seq replacement ({1..10}) [duplicate]

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})

Solution 1:

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

i: 1
i: 2
i: 3
i: 4
i: 5
i: 6
i: 7
i: 8
i: 9
i: 10

You should consider using seq(1). You can also use eval:

eval echo {$start..$end}

And here is seq

seq -s' ' $start $end

You have to use eval:

eval echo {$start..$end}

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 ""