w3resource

JavaScript: Flatten an object with the paths for keys

JavaScript fundamental (ES6 Syntax): Exercise-233 with Solution

Write a JavaScript program to flatten an object with the paths for keys.

  • Use recursion.
  • Use Object.keys(obj) combined with Array.prototype.reduce() to convert every leaf node to a flattened path node.
  • If the value of a key is an object, the function calls itself with the appropriate prefix to create the path using Object.assign().
  • Otherwise, it adds the appropriate prefixed key-value pair to the accumulator object.
  • You should always omit the second argument, prefix, unless you want every key to have a prefix.

Sample Solution:

JavaScript Code:

//#Source https://bit.ly/2neWfJ2 
const flattenObject = (obj, prefix = '') =>
  Object.keys(obj).reduce((acc, k) => {
    const pre = prefix.length ? prefix + '.' : '';
    if (typeof obj[k] === 'object') Object.assign(acc, flattenObject(obj[k], pre + k));
    else acc[pre + k] = obj[k];
    return acc;
  }, {});
console.log(flattenObject({ a: { b: { c: 1 } }, d: 1 }));

Sample Output:

{"a.b.c":1,"d":1}

Flowchart:

flowchart: Flatten an object with the paths for keys.

Live Demo:

See the Pen javascript-basic-exercise-233-1 by w3resource (@w3resource) on CodePen.


Improve this sample solution and post your code through Disqus

Previous: Write a JavaScript program that takes a function as an argument, then makes the first argument the last.
Next: Write a JavaScript program to flatten a given array up to the specified depth.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



JavaScript: Tips of the Day

How to check whether a string contains a substring in JavaScript?

ECMAScript 6 introduced String.prototype.includes:

const string = "foo";
const substring = "oo";

console.log(string.includes(substring));

includes doesn't have Internet Explorer support, though. In ECMAScript 5 or older environments, use String.prototype.indexOf, which returns -1 when a substring cannot be found:

var string = "foo";
var substring = "oo";

console.log(string.indexOf(substring) !== -1);

Ref: https://bit.ly/3fFFgZv