sed to match zero or more number of spaces in a string

Question

function getVal {
  sedPattern='s/^.*"keyVal":"\([^"]*\)".*$/\1/'
  finalSedPattern=${sedPattern/keyVal/$2}
  echo $(sed $finalSedPattern  <<< $1)
}

This is my Json parser written using sed. It takes json string, key name and returns the value like,

myJson='{"hello":"sk"}'
val=$(getVal $myJson hello)
echo $val

prints, sk

But sometimes, my json string may or may not contain space as,

myJson='{"hello" : "sk"}'

In that case, the function fails. I tried with tweaking the above pattern by adding [ ] to match zero or more spaces as,

sedPattern='s/^.*"keyVal"[ ]+:"\([^"]*\)".*$/\1/'

It throws error as,

unterminated `s' command

How can i give non-capturing pattern groups inside sed?

Solution

Your sedPattern has some issue with quotes. you are trying to match the same quote twice. Also, + is used for one or more. for 0 or more, use *.
caution: code below is untested, but should get you going.

sedPattern='s/^.*"keyVal"[ ]*:.*\(".*"\).*$/\1/'