string - Shell script - remove first and last quote (") from a variable

Question

Below is the snippet of a shell script from a larger script. It removes the quotes from the string that is held by a variable. I am doing it using sed, but is it efficient? If not, then what is the efficient way?

#!/bin/sh

opt=""html\test""
temp=`echo $opt | sed 's/.(.*)/1/' | sed 's/(.*)./1/'`
echo $temp

Answer 1

Use tr to delete ":

 echo "$opt" | tr -d '"'

Note: This removes all double quotes, not just leading and trailing.