curl json output key value to variable with jq or grep

Question

I've stored curl json output to a variable. And I want to retrieve only the description and store it in another variable.

I tried jq and grep but doesn't work.

var=`curl -i -X POST -H 'Content-Type: application/json' -d '
{
    "jsonrpc": "2.0",
    "method": "trigger.get",
    "params": {
        "filter": {"value": "1"},
        "sortfield": "lastchange",
        "limit": 20
    },
    "auth": "authstring",
    "id": 1
}' http://127.0.0.1/zabbix/api_jsonrpc.php`


echo $var

{
   "jsonrpc":"2.0",
   "result":[
      {
         "triggerid":"17169",
         "expression":"{19444}=1",
         "description":"Zabbix agent on {HOST.NAME} is unreachable for 5 minutes",
         "url":"",
         "status":"0",
         "value":"1",
         "priority":"3",
         "lastchange":"1569589239",
         "comments":"",
         "error":"",
         "templateid":"13437",
         "type":"0",
         "state":"0",
         "flags":"0",
         "recovery_mode":"0",
         "recovery_expression":"",
         "correlation_mode":"0",
         "correlation_tag":"",
         "manual_close":"0",
         "details":""
      },
      {
         "triggerid":"18123",
         "expression":"{20525}=1",
         "description":"Zabbix agent on {HOST.NAME} is unreachable for 5 minutes",
         "url":"",
         "status":"0",
         "value":"1",
         "priority":"3",
         "lastchange":"1569590703",
         "comments":"",
         "error":"",
         "templateid":"13437",
         "type":"0",
         "state":"0",
         "flags":"0",
         "recovery_mode":"0",
         "recovery_expression":"",
         "correlation_mode":"0",
         "correlation_tag":"",
         "manual_close":"0",
         "details":""
      }
   ],
   "id":1
}

echo $var | jq -r '.description'

parse error: Invalid numeric literal at line 1, column 9

Any idea what's that error means? Also how could I achieve this with grep?

Figured out how to do this with grep. So now I only need to figure out what the jq error means and how to correct it.

echo $var | grep -Po '"description":.*?[^\\]",'
"description":"Zabbix agent on {HOST.NAME} is unreachable for 5 minutes",
"description":"Zabbix agent on {HOST.NAME} is unreachable for 5 minutes",

Please include a mcve. It looks like you have some garbage at the beginning — Andy
– Andy, Commented Sep 30, 2019 at 7:12
updated the question. Those are the only step. one is var=curl output and then echo $var — user630702
– user630702, Commented Sep 30, 2019 at 7:16
@JeffMercado I upvoted your comment but it actually doesn't change anything for jq since the stripped whitespaces are insignificant in JSON. It's also weird that when OP does echo $var without quotes the output seems to have linefeeds. — Aaron
– Aaron, Commented Sep 30, 2019 at 9:12

Cuong Nguyen · Accepted Answer · 2019-11-18 09:33:57Z

1

Your stdout of $var might not in a correct json format, let validate your $var at first (https://jsonlint.com/)

Then retry with following:

echo $var | jq '.result[].description'

answered Nov 18, 2019 at 9:33

Cuong Nguyen

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Comments

peak · Accepted Answer · 2019-09-30 09:15:48Z

0

Using jq, you basically have two choices:

echo $var | jq '.. | .description?'

or if you want to be more specific:

echo $var | jq '.result[] | .description?'

But there are other variations ...

answered Sep 30, 2019 at 9:15

peak

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1 Comment

user630702 Over a year ago

getting the following error echo $var | jq '.result[] | .description?' jq: error (at <stdin>:1): Cannot index string with string "result"

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