Extracting a json value from bash shell using awk without having to resort to both awk and sed

Question

The aws command returns a json string to stdout inside a bash shell.

$ aws ssm get-parameter --name /mysite/dev/email
{
    "Parameter": {
        "Name": "/mysite/dev/email",
        "Type": "String",
        "Value": "admin@example.com",
        "Version": 1
    }
}

I want to return the value of "Value" without the quotes wrapping it. I used a combination of awk and sed linux programs. Basically, awk searches each line in the input and allows you to return a line that has a matching value. Since awk scans a file line by line, allowing you to split the line into fields and perform action on matched pattern, I did this first:

$ aws ssm get-parameter --name /mysite/dev/email | awk -F: '/Value/ {print $2}'

"admin@example.com",

As you can see, I used : as a delimited via the F flag. I matched the line with the string "Value" and then printed the second part of the delimited string, which is the value "user@example.com",

But the quotes and comma are unwelcomed. So I had to use sed with the -E regex flag to replace the quotes and comma:

$ aws ssm get-parameter --name /mysite/dev/email | awk -F: '/Value/ {print $2}' | sed -E 's/"|",//g'
 admin@example.com

I got the desired result, but I would prefer to just use awk, instead of having to pipe awk to sed. Is it possible?

Answer 1

For a one-liner, you can just remove all commas and quotes first.

awk '/Value/ { gsub(/[",]/,""); print $2}'

A better translation of your awk | sed pipeline would be

awk '/Value/ { gsub(/[",]/,"",$2); print $2}'

to just alter the values in the second field.

Answer 2

Using jq to parse out the value of the Value key in the top-level Parameter structure:

aws ssm get-parameter --name /mysite/dev/email | jq -r '.Parameter.Value'

This will additionally decode any JSON encoding of the value, so that a value that is stored in JSON as \"The Admin\" <admin@example.com> comes back as "The Admin" <admin@example.com>.