Get sequence of numbers in bash without using seq

Question

I want to loop over a sequence of numbers in a BASH script, but the bound values are not constants. I know this syntax:

for year in {2000..2010}; do
  echo ${year}
done

But the values of 2000 and 2010 are changing in my case, so what I am doing right now is this:

for year in `seq ${yeari} ${yeare}`; do
  echo ${year}
done

Is there a bash-native way to do the same, without using seq?

score 0 · Accepted Answer · answered Nov 23 '18 at 08:51

Have a look at man bash:

 for (( expr1 ; expr2 ; expr3 )) ; do list ; done
          First, the arithmetic expression expr1 is evaluated according to
          the rules described  below  under  ARITHMETIC  EVALUATION.   The
          arithmetic  expression  expr2 is then evaluated repeatedly until
          it evaluates to zero.  Each time expr2 evaluates to  a  non-zero
          value,  list  is executed and the arithmetic expression expr3 is
          evaluated.  If any expression is omitted, it behaves  as  if  it
          evaluates to 1.  The return value is the exit status of the last
          command in list that is executed, or false if any of the expres-
          sions is invalid.

for ((i=yeari; i<yeare; i++))
do
  echo $i
done

You might possibly also be interested in 'Arithmetic Expansion' ($((expression))) of which you can find more in the man page.

Get sequence of numbers in bash without using seq

1 Answers1