how to set variable for bash loop array

Question

I would like to have a bash loop function like below, with decreasing sequency:

for i in {8..2}
do
...
done

And the 8 and 2 can be set as a variable, like:

start=$1
end=$2

for i in {$start..$end}
do
...
done

But seem this dose not work. How can I do this?

Thanks for all the quick answers, later I found the answer here. descending loop with variable bash

solution:

start=$1
end=$2

for i in `seq $start -1 $end`
do
...
done

Thanks~

Thanks, but one difference, I need a decrease seq as {8..2},but not {2..8}, and the `seq $start $end` dose not work. — zhihong, Aug 01 '13 at 16:23

score 1 · Answer 1 · answered Aug 02 '13 at 08:06

1

$ start=8; end=2; for ((i = start; i >= end; i--)); do echo "${i}"; done
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answered Aug 02 '13 at 08:06

Adrian Frühwirth

score 0 · Answer 2 · answered Aug 01 '13 at 15:00

0

Nope. But here is a workaround

start=$1
end=$2

for i in $(seq $start $end)
do
...
done

answered Aug 01 '13 at 15:00

the.malkolm

score 0 · Answer 3 · answered Aug 01 '13 at 15:01

You can't use variable substitution there since the {n..m} is already one. Try using seq:

for i in `seq $start $end`
do
  ...
done

Alternatively you could do a while loop incrementing the loop variable by manually:

i=$start
while [ $i -lt $end ]; do
  ...
  : $[i++]
done

Although with while you have to be aware if $start is smaller or greater than $end

3 Answers3