Bash for loop on variable numbers

Question

I have a situation where I have large number of numbered variables. I want to evaluate each variable and set variable to a specific string if the condition is matched.

#!/bin/bash
var1=""
var2="1233123213"
var3="22332323222324242"
var4=""
var5=""

for i in 1 2 3 4 5
do 
   if [ -z "$var{$}i" ]
      then
          var{$}i="None"
   fi
echo "var{$}i \r"
done

but the problem is when I run the script I get following.

{1} \r
{2} \r
{3} \r
{4} \r
{5} \r

How I can fix this.

Inian · Accepted Answer · 2017-02-06 11:47:01Z

Use indirect variable expansion in bash with syntax {!var}.

From the man bash page,

If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exclamation point must immediately follow the left brace in order to introduce indirection.

Modify your code to something like below,

for i in 1 2 3 4 5
do
     var="var$i"
     [ -z "${!var}" ] && declare "var$i"="none"
done

printf "var1=%s\n" "$var1"
printf "var2=%s\n" "$var2"
printf "var3=%s\n" "$var3"
printf "var4=%s\n" "$var4"
printf "var5=%s\n" "$var5"

The syntax "${!var}" in this case evaluates the value of the variable within the string var which is var1, var2, var3... and the declare syntax sets the variable value at run-time, only for those variables that are empty. Now on printing those variables produces,

var1=none
var2=1233123213
var3=22332323222324242
var4=none
var5=none

Fred · Accepted Answer · 2017-02-06 11:48:46Z

1

Indirect assignment will work here, but in this specific case arrays seem like a good fit :

#!/bin/bash
declare -a var=()
var+=("")
var+=(1233123213)
var+=(22332323222324242)
var+=("")
var+=("")

for i in "${!var[@]}"
do
  [[ "${var[$i]}" ]] || var[$i]="None"
  echo "Index: $i - Value: ${var[$i]}"
done

answered Feb 6, 2017 at 11:48

Fred

7,09512 silver badges22 bronze badges

6 Comments

Inian Over a year ago

Appreciate your logic here, but this is changing OP's requirement to use dynamic variable names.

Fred Over a year ago

This uses a numerical-indexed array, but you could use an associative array for more complex cases. Arrays are a form of indirect assignment that, when it works, is often going to be cleaner/safer than initializing many separate variables.

Inian Over a year ago

But I assume, if OP prefers a way, it is better to stick that way, than to deviate with other possible ways

Fred Over a year ago

@Inian Maybe the OP really wants/needs separate variables, maybe the OP simply does not know how to use arrays to achieve the desired result, in which case not answering the question exactly as asked might also be useful.

Inian Over a year ago

It is not safe to assume OP doesn't know how to use arrays, it is good to recommend of safe alternate practices if their own logic will not help solve the problem. Who are we to decide if using dynamic variable assignment is only a small part of something big OP is trying to do !

|

MarcM · Accepted Answer · 2017-02-06 11:52:29Z

1

Consider using an array instead of numbered variables:

#!/bin/bash
var[1]=""
var[2]="1233123213"
var[3]="22332323222324242"
var[4]=""
var[5]=""

for i in 1 2 3 4 5
do 
   if [ -z "${var[i]}" ]
      then
          var[i]="None"
   fi
echo "${var[i]} \r"
done

answered Feb 6, 2017 at 11:52

MarcM

2,2511 gold badge25 silver badges32 bronze badges

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