Bash seq to produce two separate sequences of numbers in a for loop

Question

I would like to create a simple for bash loop that iterates over a sequence of numbers with a specific interval and then a different sequence of numbers, e.g.

for i in $(seq 0 5 15)
do
    echo $i
done

But after interating through i=0, 5, 10, 15, I'd like it to iterate through say 30, 35, 40, 45 as well.

Is there a way to do this using seq? Or an alternative?

Answer 1

Approach 1

Simply augment the command within $(...) with another call to seq:

for i in $(seq 0 5 15; seq 30 5 45); do
    echo $i
done

and then

$ bash test.sh
0
5
10
15
30
35
40
45

# Approach 2

In your follow-up comment, you write

The actual content of the for loop is more than just echo$i (about 200 lines) I don't want to repeat it and make my script huge!

As an alternative to the approach outlined above, you could define a shell function for those 200 lines and then call the function in a series of for loops:

f() {
   # 200 lines
   echo $i
}

for i in $(seq 0 5 15) do
    f
done

for i in $(seq 30 5 45) do
    f
done

Approach 3 (POSIX-compliant)

For maximum portability across shells, you should make your script POSIX-compliant. In that case, you need have to eschew seq, because, although many distributions provide that utility, it's not defined by POSIX.

Since you can't use seq to generate the sequence of integers to iterate over, and because POSIX doesn't define numeric, C-style for loops, you have to resort to a while loop instead. To avoid duplicating the code related to that loop, you can define another function (called custom_for_loop below):

custom_for_loop () {
  # $1: initial value
  # $2: increment
  # $3: upper bound
  # $4: name of a function that takes one parameter
  local i=$1
  while [ $i -le $3 ]; do
      $4 $i
      i=$((i+$2))
  done
}

f () {
    printf "%s\n" "$1"
}

custom_for_loop 0 5 15 f
custom_for_loop 30 5 45 f