Operate on variables in Bash directly in a loop

My googlefu is failing me. I have a bunch of variables I read it from a CSV where I want to strip whitespace. I could

variable1="$(echo -e "${variable1}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
variable2="$(echo -e "${variable2}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
variable3="$(echo -e "${variable3}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"

but then I'm copying in and editing this for every variable change. It seems to me, from other languages, there should be a way to do a simple for loop that applies to the variables as named and strips the whitespace, but I can't figure out exactly how to do that.

for $i in variable1 variable2 variable3
do
$i=$(echo -e "${$i}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
done

but this doesn't work as far as I can tell. What I want to do is loop over all the variables, strip whitespace, but not have to manually type out each variable name in the script.

Solution

Try the following:

# Sample values that need trimming.
variable1='  a  '
variable2='      b  c  '
variable3='        sadf sdf sdf     '

for name in ${!variable*}; do
  # Trim leading and trailing whitespace.
  read -r $name <<<"${!name}"
  # Print result
  echo "Variable $name now contains: [${!name}]"
done

The solution relies on variable indirection, the ability to refer to a variable indirectly, via another variable containing the target variable's name.

^{Note that variable indirection has its pitfalls (see link) and that there often better solutions, e.g., through use of arrays.}

${!variable*} expands to the names of all defined variables whose name starts with variable.
${!name} expands to the value of the variable whose name is stored in variable $name.
read -r, by virtue of reading into a single variable, implicitly trims leading and trailing whitespace.
- -r ensures that \ chars. in the input are left alone, which is typically what you want.