CODEWITHSUNDEEP
bashvariablesunixwhile-loop
NewEffect
NewEffect

Reputation: 93

While loop creating multiply variables with counter

#!/bin/bash

n=1
while (( $n <= 5 ))
do
  num$n=`echo "$n"`
  n=$(( n+1 ))

done
echo "$num1"

ok so what I am trying to do is create a while loop that will create variables and just put something into it in this case its just the value of n but i cant get it to do this!

so basically it will create num1, num2, num3 and so on

echo "$num1"
echo "$num2"
echo "$num3"

should display 

1
2
3

but i keep getting an error am i missing something here cause it shouldnt be anything crazy to do this...

Upvotes: 0

Views: 419

Answers (2)

piarston
piarston

Reputation: 1865

Try with 

#!/bin/bash

    n=1
    while (( $n <= 5 ))
    do
      eval num$n=`echo "$n"`
      n=$(( n+1 ))

    done
    echo "$num1"
    echo "$num2"
    echo "$num3"

The problem here is that bash is trying to evaluate num$n as a command, which does not exist, so the error.

Upvotes: 3

chepner
chepner

Reputation: 531095

Don't dynamically create numbered variable names like this; use an array.

n=1
while (( $n <= 5 )); do
    nums[$n]=$n  # No need to execute $(echo ...)
    n=$((n+1))
done

echo "${num[1]}"
echo "${num[2]}"
echo "${num[3]}"

Upvotes: 0

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