JavaScript : Extract properties from JSON object

Question

Problem statement : I am having a JSON object contains n number of properties inside it. I need to pass some properties from this JSON object as a JSON string to the server.

Tried : I used Object.defineProperty() method to make the enumerable as false of the object properties that I don't want to pass in my JSON string. Please find below the code snippet to get more clear picture.

var jsonObj = {
  "name": "Rohit",
  "age": 27,
  "city": "Gurgaon"
};

Object.defineProperty(jsonObj, 'name', {
  enumerable: false
});

console.log(JSON.stringify(jsonObj));

Problem facing with above code :

Suppose i have a 100 number of properties in an object and i want to pass only 20 properties to the server out of 100. Then i have to write below line of code 80 times to remove the unwanted properties and it will affect the performance of the application.

Object.defineProperty(jsonObj, 'name', { enumerable: false });

I also looked into already asked questions on SO but did not find any suitable answer for this problem statement.

Is there any better way to achieve this ?

How do you decide which properties to send to the server? Do you have an array of property names? — 31piy
– 31piy, Commented Jun 10, 2018 at 7:19
Did you compare if it wouldn't be faster to send the whole object to the server? — baao
– baao, Commented Jun 10, 2018 at 7:21

Felix Kling · Accepted Answer · 2018-06-10 07:38:45Z

3

There is always the option to create a new object and only pull out the keys you need. See One-liner to take some properties from object in ES 6 for that.

JSON.stringify accepts a "replacer" function as second argument. That allows you to change values during serialization, including filtering them out. Assuming there are fewer properties that you want to keep, you should list those in an array or Set.

var jsonObj = {
  "name": "Rohit",
  "age": 27,
  "city": "Gurgaon"
};
var toKeep = ['age', 'city']; 

console.log(JSON.stringify(
  jsonObj,
  (key, value) => !key || toKeep.indexOf(key) > -1 ? value : undefined,
));

edited Jun 10, 2018 at 7:38

answered Jun 10, 2018 at 7:31

Felix Kling

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Comments

zemil · Accepted Answer · 2018-06-10 15:46:27Z

1

You should create a new object you want to send to server

My solution:

var jsonObj = {
  "name": "Rohit",
  "age": 27,
  "city": "Gurgaon"
};

// if you don't need specific keys you should delete this and ~keys.indexOf(prop) in condition
var keys = ['name', 'city', 'moscow'];

function newObj(obj, count) {
  var newObj = {};
  var counter = count;

  for (var prop in obj) {
    if (jsonObj.hasOwnProperty(prop) && ~keys.indexOf(prop)) {
      newObj[prop] = obj[prop];
      counter -= 1;
    }

    if(counter === 0) break;
  }

  return newObj;
}

console.log(JSON.stringify(newObj(jsonObj, 20)));

edited Jun 10, 2018 at 15:46

answered Jun 10, 2018 at 7:34

zemil

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4 Comments

Felix Kling Over a year ago

for...in can iterate over properties in arbitrary order, so this would pick 20 random properties?

zemil Over a year ago

@FelixKling, I dont understand your question

Felix Kling Over a year ago

I guess I should turn it into a statement: this solution will select 20 arbitrary properties, which doesn’t seem to be what the OP wants.

zemil Over a year ago

@FelixKling, And now? I changed the code :) do you like?

Ullas Hunka · Accepted Answer · 2018-06-10 07:17:28Z

0

var jsonObj = {
  "name": "Rohit",
  "age": 27,
  "city": "Gurgaon"
};

var newObj = jsonObj;
var arr = ["name","age"];

for(var i=0;i<arr.length;i++){
  delete newObj[arr[i]]
}

console.log(newObj);

This may solve your problem I feel.

edited Jun 10, 2018 at 7:17

answered Jun 10, 2018 at 7:04

Ullas Hunka

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Comments

amrender singh · Accepted Answer · 2018-06-10 07:24:36Z

0

Try the following , It will not modify your original object:

var jsonObj = {
  "name": "Rohit",
  "age": 27,
  "city": "Gurgaon"
};
var storePropeties = ["city"];
var result = Object.keys(jsonObj).reduce(function (obj, key) {
   if(storePropeties.includes(key))
      obj[key] = jsonObj[key];
     return obj
 }, {})

console.log(result);

answered Jun 10, 2018 at 7:24

amrender singh

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