Concatenate two JSON objects

Based on your description in the comments, you'd simply do an array concat:

var jsonArray1 = [{'name': "doug", 'id':5}, {'name': "dofug", 'id':23}];
var jsonArray2 = [{'name': "goud", 'id':1}, {'name': "doaaug", 'id':52}];
jsonArray1 = jsonArray1.concat(jsonArray2);
// jsonArray1 = [{'name': "doug", 'id':5}, {'name': "dofug", 'id':23}, 
//{'name': "goud", 'id':1}, {'name': "doaaug", 'id':52}];

If you'd rather copy the properties:

var json1 = { value1: '1', value2: '2' };
var json2 = { value2: '4', value3: '3' };


function jsonConcat(o1, o2) {
 for (var key in o2) {
  o1[key] = o2[key];
 }
 return o1;
}

var output = {};
output = jsonConcat(output, json1);
output = jsonConcat(output, json2);

Output of above code is{ value1: '1', value2: '4', value3: '3' }

I use:

let x = { a: 1, b: 2, c: 3 }

let y = {c: 4, d: 5, e: 6 }

let z = Object.assign(x, y)

console.log(z)

// OUTPUTS:
{ a:1, b:2, c:4, d:5, e:6 }

From here.

One solution is to use a list/array:

var first_json = {"name":"joe", "age":27};
var second_json = {"name":"james", "age":32};

var jsons = new Array();
jsons.push(first_json);
jsons.push(second_json);

Result

jsons = [
    {"name":"joe", "age":27},
    {"name":"james", "age":32}
]

if using TypeScript, you can use the spread operator (...)

var json = {...json1,...json2} 

You can use Object.assign() method. The Object.assign() method is used to copy the values of all enumerable own properties from one or more source objects to a target object. It will return the target object.[1]

var o1 = { a: 1 }, o2 = { b: 2 }, o3 = { c: 3 };

var obj = Object.assign(o1, o2, o3);
console.log(obj); // { a: 1, b: 2, c: 3 }

okay, you can do this in one line of code. you'll need json2.js for this (you probably already have.). the two json objects here are unparsed strings.

json1 = '[{"foo":"bar"},{"bar":"foo"},{"name":"craig"}]';

json2 = '[{"foo":"baz"},{"bar":"fob"},{"name":"george"}]';

concattedjson = JSON.stringify(JSON.parse(json1).concat(JSON.parse(json2)));

var baseArrayOfJsonObjects = [{},{}];
for (var i=0; i<arrayOfJsonObjectsFromAjax.length; i++) {
    baseArrayOfJsonObjects.push(arrayOfJsonObjectsFromAjax[i]);
}

I use:

let jsonFile = {};    
let schemaJson = {};    
schemaJson["properties"] = {};    
schemaJson["properties"]["key"] = "value";
jsonFile.concat(schemaJson);

The simplest way :

const json1 = { value1: '1', value2: '2' };
const json2 = { value2: '4', value3: '3' };

const combinedData = {
  json1,
  json2
};

console.log(combinedData)

I dont know if you want this:

U can use this for create from arrays, all arrays need contains the same number of elments.

Example: If you have:

let a = ["a", "b", "c"];
let b = [1, 2, 3];

Use

concatArraysLikeJson([a, b]);

The result of is:

let result = {
 0 : ["a", 1],
 1 : ["b", 2],
 2 : ["c", 3]
};

Typescript

concatArraysLikeJson(arrays:any){
 let result:any = {};

 let size:number = 0;
 let make:boolean = true;

 if(arrays.length > 0){
  size = arrays[0].length;

  for(let i = 1; i < arrays.length; i++){
    let array = arrays[i];
    
    if(make){
      if(array.length != size){
        make = false;
      }
    }
  }
 }

 if(make){
  for (let o = 0; o < size; o++) {
    result[o] = [];
  }

  for(let i = 0; i < arrays.length; i++){
    const array = arrays[i];

    //console.log(array);

    for (let o = 0; o < size; o++) {
      const element = array[o];
      result[o].push(element);
    }
  }

  return result;
 }else{
  return false;
 }
}

Javascript:

concatArraysLikeJson(arrays){
 let result = {};

 let size = 0;
 let make = true;

 if(arrays.length > 0){
  size = arrays[0].length;

  for(let i = 1; i < arrays.length; i++){
    let array = arrays[i];
    
    if(make){
      if(array.length != size){
        make = false;
      }
    }
  }
}

if(make){
  for (let o = 0; o < size; o++) {
    result[o] = [];
  }

  for(let i = 0; i < arrays.length; i++){
    const array = arrays[i];

    //console.log(array);

    for (let o = 0; o < size; o++) {
      const element = array[o];
      result[o].push(element);
    }
  }

  return result;
}else{
  return false;
}
  }