Assigning Dynamic Bash Variable Names Using a for Loop Seq

Assigning dynamic bash variable names using a for loop seq

Kurt Stutsman provides the right pointer in a comment on the question: use Bash arrays to solve your problem.

Here's a simplified example:

groups=() # declare an empty array; same as: declare -a groups
for i in {0..5}; do
  groups[i]="group $i"  # dynamically create element with index $i
done

# Print the resulting array's elements.
printf '%s\n' "${groups[@]}"

^{See the bottom of this answer for other ways to enumerate the elements of array ${groups[@]}.}

• bash arrays can be dynamically expanded (and can even be sparse - element indices need not be contiguous)

  • Hence, simply assigning to element $i works, without prior sizing of the array.

• Note how $i need not be prefixed with $ in the array subscript, because array subscripts are evaluated in an arithmetic context (the same context in which $(( ... )) expressions are evaluated).

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As for what you did wrong:

group$i=...

is not recognized as a variable assignment by Bash, because - taken literally - group$i is not a valid identifier (variable name).

Because it isn't, Bash continues to parse until the next shell metacharacter is found, and then interprets the resulting word as a command to execute, which in your case resulted in error message group0=j: command not found.

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If, for some reason, you don't want to use arrays to avoid this problem entirely, you can work around the problem:

By involving a variable-declaring builtin [command] such as declare, local, or export, you force Bash to perform expansions first, which expands group$i to a valid variable name before passing it to the builtin.

• user2683246's answer demonstrates the next best approach by using declare (or, if local variables inside a function are desired, local) to create the variables.

• Soren's answer uses export, but that is only advisable if you want to create environment variables visible to child processes rather than mere shell variables.

Caveat: With this technique, be sure to double-quote the RHS in order to capture the full value; to illustrate:

 i=0; declare v$i=$(echo 'hi, there'); echo "$v0" # !! WRONG -> 'hi,': only UP TO 1ST SPACE

 i=0; declare v$i="$(echo 'hi, there')"; echo "$v0" # OK -> 'hi, there'

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Other ways to enumerate the groups array created above:

# Enumerate array elements directly.
for element in "${groups[@]}"; do
  echo "$element"
done

# Enumerate array elements by index.
for (( i = 0; i < ${#groups[@]}; i++ )); do
  echo "#$i: ${groups[i]}"
done

Dynamic variable names in Bash

Use an associative array, with command names as keys.

# Requires bash 4, though
declare -A magic_variable=()

function grep_search() {
    magic_variable[$1]=$( ls | tail -1 )
    echo ${magic_variable[$1]}
}

If you can't use associative arrays (e.g., you must support bash 3), you can use declare to create dynamic variable names:

declare "magic_variable_$1=$(ls | tail -1)"

and use indirect parameter expansion to access the value.

var="magic_variable_$1"
echo "${!var}"

See BashFAQ: Indirection - Evaluating indirect/reference variables.

Bash declaring variable with a number inside a for loop

Without an array, you need to use the declare command:

i=0
for int in $ints; do
    i=$((i +1))
    declare "intf$i=$int"
done

With an array:

intf=()
for int in $ints; do
    intf+=( $int )
done

Increment variable name results in command not found

To achieve the outcome required, the use of arrays are needed and so:

#!/bin/bash
for i in $(seq 5)
do
  var[$i]="number is $i"
done
for i in "${var[@]}"
do
    echo "$i"
done

Set the index and values for the array var accordingly and then loop through the array and print the values.

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