How can I toggle an interface every second in a Bash loop?

Question

I want to put an interface down and up every 1 second for 80 times, how can I implement this by a bash script?

Something like this?

COUNT = 80
for n in $(seq -w 1 $COUNT); do
    case $n in   
    [1,3,5,7,9....79]*) # I don't know how to represent the odd value only
       ifconfig veth1 down
       sleep 1
       ;;
    [2,4,6,8,10....80]*)
       ifconfig veth1 up
       sleep 1
       ;;
   esac
done 

Answer 1

COUNT=40
for n in $(seq -w 1 $COUNT); do
  ifconfig veth1 down
  sleep 1
  ifconfig veth1 up
  sleep 1
done

Or if you really want to count to 80:

COUNT=80
for n in $(seq -w 1 $COUNT); do
  case $n in
    *[13579])
     ifconfig veth1 down
     ;;
    *)
     ifconfig veth1 up
     ;;
  esac
  sleep 1
done

Answer 2

Toggle a flag:

#!/bin/bash
for ((i = 1, flag = 0; i <= 80; i++))
do
    if ((flag ^= 1))
    then
        ifconfig veth1 down    # odd
    else
        ifconfig veth1 up
    fi
    sleep 1
done

Answer 3

use % operator. like the following, replace the echo with the commands you want

count=0
while [ $count -lt 80 ]
do
    if (( $count % 2 == 0 ))
    then
        echo 'aaa'
    else
        echo 'bbb'
    fi

    count=$(( $count + 1 ))
done

Answer 4

If you don't mind the bashisms, you can make your code much more concise through the use of various expansions available in Bash. For example:

for i in {1..80}; do
    case $((i % 2)) in
        0) ifconfig veth1 down ;;
        1) ifconfig veth1 up   ;;
    esac
    sleep 1
done

The magic here is the {1..80} sequence expression, coupled with the modulo operator to determine whether the number is odd or even. If your version of Bash doesn't support sequence expressions for any reason, just use $(seq 1 80) instead.