12

Is there a way to create bash variables and assign them values via a loop?

Something along the lines of:

#!/bin/bash

c=0
for file in $( ls ); do
    var"$c"="$file";
    let c=$c+1;
done

EDIT: Thank you to @Costas and @mdpc for pointing out this would be a poor alternative to a list; the question is theoretical only.

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  • 1
    NOTE: Be aware that your setup will change/modify/assign variables only in the current process and NOT the parent process if you execute this as a command. You could source the file using the '.' command (EX: . new_vars) to make this happen within the current process. – mdpc Apr 18 '15 at 20:29
  • Ok, but the above doesn't work actually run; var"$c" breaks it. @Costas 's answer demonstrates how to assign multiple values to a single list variable - I'm just asking theoretically whether variables can actually be generated with a loop. – toxefa Apr 18 '15 at 20:34
  • Hmmm...why so complicated, just using a new variable creates it without any fanfare. – mdpc Apr 18 '15 at 20:36
14

Well, you absolutely can using eval as follows:

c=0
for file in $( ls ); do
    eval "var$c=$file";
    c=$((c+1));
done

This code will create variables named var0, var1, var2, ... with each one holding the file name. I assume you will have a good reason you want to do that over using an array ...

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5

Possible you mean array. There are some ways to assign values

First:

c=0
for file in $( ls ); do
    var[$c]="$file";
    c=$(($c+1));
done

Second:

c=0
for file in $( ls ); do
    var[c++]="$file";
done

Third:

for file in $( ls ); do
    var[${#var}]="$file";
done

Fourth:

var=( $(ls) )

Fifth

...

  1. there is no need to use ls , just put * for all files
  2. inside [] the $ sign can be ommited
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  • Thank you; my question is potentially more theoretical than actually practical. However, when I echo $var with the commands you list above I only get the first file 'listed' by ls. Am I doing something wrong? – toxefa Apr 18 '15 at 20:19
  • 1
    If you wants to output all array you have to use echo ${var[*]} (or ${var[@]}), single element can be called like echo ${var[1]}. Note: $var is equal ${var[0]} – Costas Apr 18 '15 at 20:25
  • 1
    @py4on Other way is use a loop: for v in "${var[@]}";do echo $v;done – Costas Apr 18 '15 at 20:29

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