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

Use tr to delete ":

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

NOTE: This does not fully answer the question, removes all double quotes, not just leading and trailing. See other answers below.

If you're using jq and trying to remove the quotes from the result, the other answers will work, but there's a better way. By using the -r option, you can output the result with no quotes.

$ echo '{"foo": "bar"}' | jq '.foo'
"bar"

$ echo '{"foo": "bar"}' | jq -r '.foo'
bar

There is a straightforward way using xargs:

> echo '"quoted"' | xargs
quoted

xargs uses echo as the default command if no command is provided and strips quotes from the input, see e.g. here. Note, however, that this will work only if the string does not contain additional quotes. In that case it will either fail (uneven number of quotes) or remove all of them.

There's a simpler and more efficient way, using the native shell prefix/suffix removal feature:

temp="${opt%\"}"
temp="${temp#\"}"
echo "$temp"

${opt%\"} will remove the suffix " (escaped with a backslash to prevent shell interpretation).

${temp#\"} will remove the prefix " (escaped with a backslash to prevent shell interpretation).

Another advantage is that it will remove surrounding quotes only if there are surrounding quotes.

BTW, your solution always removes the first and last character, whatever they may be (of course, I'm sure you know your data, but it's always better to be sure of what you're removing).

Using sed:

echo "$opt" | sed -e 's/^"//' -e 's/"$//'

(Improved version, as indicated by jfgagne, getting rid of echo)

sed -e 's/^"//' -e 's/"$//' <<<"$opt"

So it replaces a leading " with nothing, and a trailing " with nothing too. In the same invocation (there isn't any need to pipe and start another sed. Using -e you can have multiple text processing).