'Grep output based on regex: multiline with space bash

I have the following output from my command. I am trying to extract the id from the text if possible as a one liner

# module.mongodb_atlas.mongodbatlas_project_ip_whitelist.vpc-whitelist["10.3.0.3/17"]:
resource "mongodbatlas_project_ip_whitelist" "vpc-whitelist" {
    cidr_block = "10.3.0.0/17"
    comment    = "External access from '10.3.0.3/17' to the MongoDB Atlas"
    id         = "dffgsgsnk=:sdgsdg=-cHJvamVjdF9pZA==:sdfsdfs"
    project_id = "sdsdss"

    timeouts {}
}

I am trying to get the id from the above text using bash.

I tried converting the output to json but it did not work as the output can have different formats which would make it difficult.

echo $json| tail -n +3 | sed 's/[[:blank:]]//g' i removed the empty space. But the sed part of extracting the id= hasnt worked.

echo $json| tail -n +3 | sed 's/[[:blank:]]//g' | sed -e '/id=/!d'

I also tried different ways of grep to get the id but nothing worked. Everything gave an empty string. grepwith id would give me 3 lines as output so i need id= to be the matching string

echo $json| tail -n +3 | sed 's/[[:blank:]]//g' | grep 'project_id'

project_id="dfgdfg"

echo $json| tail -n +3 | sed 's/[[:blank:]]//g' | grep '="' (Once i add id=" here nothing comes up)

cidr_block="10.3.0.3/17" comment="Externalaccessfrom'10.3.0.3/17'totheMongoDBAtlas" id="ZW50cnk=:fdgdfg=-dfg==:dg" project_id="dfgdfg"

bashsedgrep


Solution 1:[1]

Using sed

$ sed -n '/\<id\>/ {s/[^"]*"\([^"]*\).*/\1/p}' <<< "$json"
dffgsgsnk=:sdgsdg=-cHJvamVjdF9pZA==:sdfsdfs

Solution 2:[2]

using awk, without converting it to json:

awk '/\yid\y/ { gsub(/"/, "");print $3}' file
  • /\yid\y/ exact match of the id word
  • gsub(/"/, "") quotes removal
  • print $3 print the id field

Sources

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Source: Stack Overflow

Solution Source
Solution 1 HatLess
Solution 2

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