You need to enclose in brackets and double quotes: jq ‘.”component-status”‘ With your given input it returns: [ { “status”: “OK”, “component”: “Service1”, “status-code”: 200 }, { “status”: “OK”, “component”: “Service2”, “status-code”: 200 } ] The jq Manual (development) –> Basic filters: .foo, .foo.bar The simplest useful filter is .foo. When given a JSON object … Read more
Adapted from this post on Processing JSON with jq, you can use the select(bool) like this: $ jq ‘.[] | select(.location==”Stockholm”)’ json { “location”: “Stockholm”, “name”: “Walt” } { “location”: “Stockholm”, “name”: “Donald” }
The |= .+ part in the filter adds a new element to the existing array. You can use jq with filter like: jq ‘.data.messages[3] |= . + {“date”: “2010-01-07T19:55:99.999Z”, “xml”: “xml_samplesheet_2017_01_07_run_09.xml”, “status”: “OKKK”, “message”: “metadata loaded into iRODS successfullyyyyy”}’ inputJson To avoid using the hardcoded length value 3 and dynamically add a new element, use … Read more
Since 1.4 this is now possible with the * operator. When given two objects, it will merge them recursively. For example, jq -s ‘.[0] * .[1]’ file1 file2 Important: Note the -s (–slurp) flag, which puts files in the same array. Would get you: { “value1”: 200, “timestamp”: 1382461861, “value”: { “aaa”: { “value1”: “v1”, … Read more
First, obtain an array containing all the different object property names in your object array input. Those will be the columns of your CSV: (map(keys) | add | unique) as $cols Then, for each object in the object array input, map the column names you obtained to the corresponding properties in the object. Those will … Read more
Consider also passing in the shell variable (EMAILID) as a jq variable (here also EMAILID, for the sake of illustration): projectID=$(jq -r –arg EMAILID “$EMAILID” ‘ .resource[] | select(.username==$EMAILID) | .id’ file.json) Postscript For the record, another possibility would be to use jq’s env function for accessing environment variables. For example, consider this sequence of … Read more
You could use sort_by/1 on the fields, you’ll have to either normalize the date/time formats or conditionally parse them out using strptime/1. It returns an array of the date fields in significant order combine to a single array and sort_by will do the rest. .data.monitor | sort_by([ (.Date | strptime(“%A %d %B %Y”)), (.Time | … Read more