同时轻松增加和打印值

Question

I need help incrementing array values inside loops. The problem is variables are all the same and the second element of "Numbers" array is not incremented.

#!/bin/bash

Duration=60

declare -a Numbers=("5" "10")

for (( d=1 ; d<=$Duration ; d++ ))
do

  for (( i=0 ; i<${#Numbers[@]} ; i++ ))
  do

        if [ "$MYVALA" == "" ]; then
                MYVALA=${Numbers[i]}
        else
                MYVALA=$(($MYVALA+1))
        fi ;

        echo ""
        echo "number: ${Numbers[i]}"
        echo "-------------"
        echo "new value = $MYVALA"
  done ;
  sleep 1 ;
done ;

this is the result of code above:

number: 5
-------------
new value = 5

number: 10
-------------
new value = 6

number: 5
-------------
new value = 7

number: 10
-------------
new value = 8

What I would like to get is:

number: 5
-------------
new value = 6

number: 10
-------------
new value = 11

number: 5
-------------
new value = 7

number: 10
-------------
new value = 12
...

number 5 and number 10 are printed at the same time and once per second.

Thanks for your help.

Answer 1

This produces the output you wanted. The new value is simply the number plus duration.

#!/bin/bash

Duration=60
Numbers=(5 10)

for (( d=1 ; d<=Duration ; d++ )) ; do
  for (( i=0 ; i<${#Numbers[@]} ; i++ )) ; do
      let MYVALA=Numbers[i]+d
      echo
      echo "number: ${Numbers[i]}"
      echo '-------------'
      echo "new value = $MYVALA"
  done
  sleep 1
done

Answer 2

To increment an array value, use (( myarray[i]++ )) . To make your script print out the values you describe, you can keep a separate array of counters for each number.

#!/bin/bash

Duration=60

declare -a Numbers=("5" "10")
Counters=( "${Numbers[@]}" )

for (( d=1 ; d<=$Duration ; d++ ))
do

  for (( i=0 ; i<${#Numbers[@]} ; i++ ))
  do
        (( Counters[i]++ ))
        echo ""
        echo "number: ${Numbers[i]}"
        echo "-------------"
        echo "new value = ${Counters[i]}"
  done ;
  sleep 1 ;
done