How to extract this specific text/json only using grep

Question

So I have this json file that's an output from a previous command:

{
    "upload": {
        "status": "INITIALIZED", 
        "contentType": "application/octet-stream", 
        "name": "mobile.apk", 
        "created": 1511111178.799, 
        "url": "https://amazonaws.com/arn%358%3Aproject%3A0030e-0a929/uploads/arn%3Aaws%3??/33bgb", 
        "type": "ANDROID_APP", 
        "arn": "arn:aws:devicefarm:us-west-2:588:upload:0000-aaa-bg01-11ab11/1feebbb-1a1a"
    }
}

I want to extract the "arn" value, including the quotes, so the value:

"arn:aws:devicefarm:us-west-2:588:upload:0000-aaa-bg01-11ab11/1feebbb-1a1a"

Should be returned. I'll also need a separate command to extract the url value also, so the following needs to be returned:

"https://amazonaws.com/arn%358%3Aproject%3A0030e-0a929/uploads/arn%3Aaws%3??/33bgb"

So I need 2 separate commands for these values. I've tried all different combinations of grep and I've not got anything to work yet.

I only want to use grep for the command, my situation won't allow to install any tools.

Thank you!

Don't use text processing tools to process JSON stream. Use a proper syntax aware parser — Inian, Dec 15 '17 at 14:25
In your case you just need `awk '/arn:/{print $2; exit}' jsonFile` or `awk '/arn:/{print $2; exit}' jsonFile` — Inian, Dec 15 '17 at 14:49
grep solutions: `cat json_file | grep -oP '(?<= "url": ).+(?=,)'` and `cat json_file | grep -oP '(?<= "arn": ).+$'` — AnnieFromTaiwan, Dec 15 '17 at 14:52

Kamil · Answer 1 · 2017-12-15T14:47:40.720

There is couple of possible solutions, depending on a tool set that is available for you. Lets assume that bash variable JSON contains whole JSON structure you described in the topic.

Bash + Python:

ARN=`echo "$JSON" | python -c "import sys, json; print json.load(sys.stdin)['upload']['arn']"`
URL=`echo "$JSON" | python -c "import sys, json; print json.load(sys.stdin)['upload']['url']"`

Bash + jq:

URL=`echo "$JSON" | jq -r '.upload.url'`
ARN=`echo "$JSON" | jq -r '.upload.arn'`

Bash + grep + awk

URL=`echo "$JSON" | grep -Po '"url":.*?".*?"' | awk '{ print $2 }'`
ARN=`echo "$JSON" | grep -Po '"arn":.*?".*?"' | awk '{ print $2 }'`

Now you can do

echo "$ARN"
echo "$URL"

But as it was already mentioned by others, don't use awk, grep, sed and any other tools for text processing. Use dedicated tools for this.

How to extract this specific text/json *only* using grep

1 Answers1

How to extract this specific text/json only using grep