How to use bash variable with double and single quotes

Question

i am writing personal parser for json on bash. So, i need to make grep of line. For example, part of my json-file.

{
    "host": "127.0.0.1",
    "hhost": "127.0.0.2",
}

Of course, after trying cat json.txt | grep host cat json.txt | grep host , i received:

"host": "127.0.0.1",
"hhost": "127.0.0.1",

I found, how to find only host without hhost . I used

cat json.txt | grep '"host"'

Everything is good

"host": "127.0.0.1",

But i want to use it in bash script:

#!/bin/bash

#in a future, i want to read variable var from reading from console 
var=host
search='"$var"'
echo $search

In a result, i have:

"$var"

What i did wrong ? Can you advice me please ?

Answer 1

You could use jq , to parse the JSON file for this you will need first a proper valid json

{
  "host": "127.0.0.1",
  "hhost": "127.0.0.2"
}

Then you could just do something like:

#/bin/sh

HOST=$(jq '.host' data.json)
echo $HOST
...

Answer 2

if you put a variable inside single quotes in bash, I does not get evaluated, you should just use:

search="$var"

if you want the variable to be enclosed by quotes, just escape them like this:

search="\"$var\""