Copying properties from one object to another with a condition

Question

Lazy-me is wondering if there a better way to copy the properties in one object (source) over to another object (destination) only if the properties exist in the latter? It does not necessarily have to be using Underscore.

For example,

_.mixin({ assign: function (o, destination, source) { for (var property in source) { if (destination.hasOwnProperty(property)) { destination[property] = source[property]; } } return destination; } }); console.log( _().assign({ a: 1, b: 2, d: 3 }, { a: 4, c: 5 }) ) // a: 4, b: 2, d: 3

Possible duplicate of How to duplicate object properties in another object? — Michael Freidgeim
– Michael Freidgeim, Commented Oct 3, 2019 at 7:22

GibboK · Accepted Answer · 2016-08-15 10:42:43Z

Use Object.assign(obj1, obj2); (if the properties exist in the latter) which is native in ES6 (no underscore.js is required).

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. More info here.

Example:

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

Alternatively use undescore.js

_.extend(destination, *sources)

or

_.extendOwn(destination, *sources)

Detailated information can be found here: http://underscorejs.org/#extend

@Vohuman thanks for your comment I have edit a note in my answer for reference.

Ram · Accepted Answer · 2016-08-15 10:47:51Z

One lazy option is:

_.extend(a, _.pick(b, _.keys(a)));

_.pick filters the source object by using the .keys of the destination object and the result is used for extending the destination object.

If you don't want to modify the original objects just pass an empty object to the _.extend function.

_.extend({}, a, _.pick(b, _.keys(a)));

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