Shell script “for” loop syntax

Question

I have gotten the following to work:

for i in {2..10}
do
    echo \"output: $i\"
done

It produces a bunch of lines of output: 2

Answer 1

These all do {1..8} and should all be POSIX. They also will not break if you put a conditional continue in the loop. The canonical way:

f=
while [ $((f+=1)) -le 8 ]
do
  echo $f
done

Another way:

g=
while
  g=${g}1
  [ ${#g} -le 8 ]
do
  echo ${#g}
done

and another:

set --
while
  set $* .
  [ ${#} -le 8 ]
do
  echo ${#}
done

Answer 2

If the seq command available on your system:

for i in `seq 2 $max`
do
  echo "output: $i"
done

If not, then use poor man's seq with perl:

seq=`perl -e "\$,=' ';print 2..$max"`
for i in $seq
do
  echo "output: $i"
done

Watch those quote marks.

Answer 3

Use:

max=10
for i in `eval echo {2..$max}`
do
    echo $i
done

You need the explicit 'eval' call to reevaluate the {} after variable substitution.

Answer 4

There's more than one way to do it.

max=10
for i in `eval "echo {2..$max}"`
do
    echo "$i"
done

Answer 5

Step the loop manually:

i=0
max=10
while [ $i -lt $max ]
do
    echo "output: $i"
    true $(( i++ ))
done

If you don’t have to be totally POSIX, you can use the arithmetic for loop:

max=10
for (( i=0; i < max; i++ )); do echo "output: $i"; done

Or use jot(1) on BSD systems:

for i in $( jot 0 10 ); do echo "output: $i"; done

Answer 6

Here it worked on Mac OS X.

It includes the example of a BSD date, how to increment and decrement the date also:

for ((i=28; i>=6 ; i--));
do
    dat=`date -v-${i}d -j "+%Y%m%d"` 
    echo $dat
done