In bash I want array let say:
array=(1 2 3)
Then I need a loop for program where
x will be 1,2,3,1,2,3... (from array)
i will be unlimited 1,2,3,4,5,6.... (main loop)
My code:
array=(1 2 3)
while true ; do
((i=i+1))
#screen -dmS plot$i -d /destinatin$x
echo $i $x
sleep 1
done
I do not know how to loop array and set $x to go 1,2,3,1,2,3....
Infinite loops are generally generated using the shell built-in command : which does nothing in its singular form. So if you want to loop infinitely over the elements of a list, you can do the following:
1. The infinite nested while-for loop:
while :; do for i in "${a[@]}"; do echo "${i}"; done; done
2. using an index-reset
i=0; while :; do echo "${a[i]}"; ((i=i+1)); ((i==${#a[@]})) && i=0; done
2. using modulo calculation:
i=0; while :; do echo "${a[i]}"; (( i=(i+1) % ${#a[@]} )); done
3. the infinite for loop with modulo index
for ((i=0;;i++)); do echo "${a[i%${#a[@]}]}"; done
This code should solve your problem:
#!/bin/bash
array=(1 2 3)
i=0
count_of_elements=${#array[@]} #counting the number of array elements
while true; do
rest=$(($i%$count_of_elements)) #counting rest of the division by count of array elements
printf "${array[$rest]}," #dispay result
i=$((i+1))
done
It will be also working if you change your input array (for example if it will be array=(1 2 3 4 5).