bash - for loop different iterations

Question

I know how to do sequential for loops like:

for i in $(seq 0 63)
do
  echo $i
done

This would print 0-63 . But what if I wanted certain numbers looped like only 0, 5, 25, 43, 44, 51, 54.

I know I could do the following:

for i in $(seq 0 63)
do
  if [ "$i" -eq 0 ] || [ "$i" -eq 5 ] || [ "$i" -eq 25 ] || [ "$i" -eq 43 ] || [ "$i" -eq 44 ] || [ "$i" -eq 51 ] || [ "$i" -eq 54 ]; then
    echo $i
  fi
done

But the result will be that it still goes through those loop iterations, and is not an efficient solution.

Is there something I can use with seq to describe these certain numbers 0,5,25,43,44,51,54?

nradk · Accepted Answer · 2017-02-15 14:57:54Z

1

If you want to iterate over a known list, you can simply store the elements, space-separated, in a variable and loop over that:

nums="0 5 25 43 44 51 54";
for i in $nums;
do echo $i;
done;

answered Feb 15, 2017 at 14:57

nradk

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1 Comment

chepner Over a year ago

For the OP, seq is just generating such a hard-coded list.

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