Appending a key value pair to a json object - json

Question

This is the json object I am working with
{
"name": "John Smith",
"age": 32,
"employed": true,
"address": {
"street": "701 First Ave.",
"city": "Sunnyvale, CA 95125",
"country": "United States"
},
"children": [
{
"name": "Richard",
"age": 7
},
{
"name": "Susan",
"age": 4
},
{
"name": "James",
"age": 3
}
]
}
I want this as another key-value pair :
"collegeId": {
"eventno": "6062",
"eventdesc": "abc"
};
I tried concat but that gave me the result with || symbol and I cdnt iterate. I used spilt but that removes only commas.
concattedjson = JSON.stringify(JSON.parse(json1).concat(JSON.parse(json2)));
How do I add a key pair value to an existing json object ?
I am working in javascript.

Answer 1

This is the easiest way and it's working to me.
var testJson = {
"name": "John Smith",
"age": 32,
"employed": true,
"address": {
"street": "701 First Ave.",
"city": "Sunnyvale, CA 95125",
"country": "United States"
},
"children": [
{
"name": "Richard",
"age": 7
},
{
"name": "Susan",
"age": 4
},
{
"name": "James",
"age": 3
}
]
};
testJson.collegeId = {"eventno": "6062","eventdesc": "abc"};

Answer 2

Just convert the JSON string to an object using JSON.parse() and then add the property. If you need it back into a string, do JSON.stringify().
BTW, there's no such thing as a JSON object. There are objects, and there are JSON strings that represent those objects.

Answer 3

You need to make an object at reference "collegeId", and then for that object, make two more key value pairs there like this:
var concattedjson = JSON.parse(json1);
concattedjson["collegeId"] = {};
concattedjson["collegeId"]["eventno"] = "6062";
concattedjson["collegeId"]["eventdesc"] = "abc";
Assuming that concattedjson is your json object. If you only have a string representation you will need to parse it first before you extend it.
Edit
demo for those who think this will not work.

Answer 4

const newTestJson = JSON.parse(JSON.stringify(testJson));
newTestJson.collegeId = {"eventno": "6062","eventdesc": "abc"};
testJson = newTestJson;