Assigning dynamic bash variable names using a for

Question

So I'm trying to do something, not sure if it's possible. I have the following code:

for i in {0..5}; do
    if [[ -f ./user$i ]]; then
        group$i=$(grep -w "group" ./user0|awk '{print $2}'|perl -lape 's/\s+//sg')

What I want to do is assign a unique variable for each instance of the {0..5} so group1 group2 group3 group4 for each variable name. Then I would change ./user0 to ./user$i and create a dynamic list of variables based on my sequence. Is this possible? I get the following error when trying to execute this and I'm unsure of what I have actually done that bash doesn't like.

test.sh: line 16: group0=j: command not found

Answer 1

Try to use the export or declare function like this

for i in {0..5}; do
    if [[ -f ./user$i ]]; then
        export group$i=$(grep -w "group" ......

with declare

for i in {0..5}; do
    if [[ -f ./user$i ]]; then
        declare group$i=$(grep -w "group" ......

where export makes the value available to sub-processes, and declare just available within the same script.

Answer 2

Use declare group$i=... instead of just group$i=...

Answer 3

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).

---

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.

---

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'

---

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