What is the best method for extracting a nested json value without a key?

Is there a way to retrieve a specific value from a nested JSON object that does not have a key?

I'm attempting to access the "2019-03-19T22:57:47.972Z" value within this JSON structure:

var json = {"metaData":[{"name":"ACTION_NAME"},{"name":"SENT_RECV_TIME"}],"rows":[["SI_OA_CTPParameters","2019-03-20T06:20:45.704Z"],["SI_OA_CTPParameters","2019-03-21T06:04:08.313Z"],["SI_OA_CTPParameters","2019-03-21T06:01:14.412Z"],["SI_OA_CTPParameters","2019-03-20T06:59:54.875Z"],["SI_OA_CTPParameters","2019-03-20T20:32:50.975Z"],["SI_OA_CloudDataAddress","2019-03-19T22:57:47.972Z"],["SI_OA_CloudDataAddress","2019-03-19T22:56:52.115Z"],["SI_OA_CloudDataAddress","2019-03-19T22:54:28.196Z"] ......

Currently, I am only able to extract json.rows[0], which gives me:

["SI_OA_CTPParameters","2019-03-21T06:04:08.313Z"]

I tried using json.rows[0].[1] but that didn't work.

All I need is the second value "2019-03-21T06:04:08.313Z", how can I access it?

javascriptarraysjsonobject

Answer №1

To retrieve nested values from the JSON object, use json.rows[0][1] in your code like so:

var json = {"metaData":[{"name":"ACTION_NAME"},{"name":"SENT_RECV_TIME"}],"rows":[["SI_OA_CTPParameters","2019-03-20T06:20:45.704Z"],["SI_OA_CTPParameters","2019-03-21T06:04:08.313Z"],["SI_OA_CTPParameters","2019-03-21T06:01:14.412Z"],["SI_OA_CTPParameters","2019-03-20T06:59:54.875Z"],["SI_OA_CTPParameters","2019-03-20T20:32:50.975Z"],["SI_OA_CloudDataAddress","2019-03-19T22:57:47.972Z"],["SI_OA_CloudDataAddress","2019-03-19T22:56:52.115Z"],["SI_OA_CloudDataAddress","2019-03-19T22:54:28.196Z"]}};

console.log(json.rows[0][1]);

Answer №2

json.rows[0] retrieves an array, which we will refer to as 'a'. Individual elements of an array can be accessed by their index, like so: a[1] will provide the element you are looking for.

Although it may not be convenient to rename the array, you can simply reinsert the original statement within 'a'; therefore, json.rows[0][1] will also work effectively.

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