How to combine object properties in typescript?

Question

I'd like to know the best way to do this, say I have two objects

var objectA = {     propertyA: 1,     propertyB: 2     ...     propertyM: 13 }  var objectB = {     propertyN: 14,     propertyO: 15     ...     propertyZ: 26 } 

If objectC is created by

var objectC = Object.assign(objectA, objectB); 

How can I declare/describe objectC, so the compiler/IDE knows that it has the properties of both objectA and objectB?

I'd like to find a way without the need of defining interfaces for objectA and objectB. I don't want to write declaration and definition/evaluation for the same property twice. This redundancy is significant if I have too many properties on an object.

(Is there an operator that can extract the interface/type of an existing object?)

Is it possible?

Answer 1

Seems like this should do the trick:

var objectA = {     propertyA: 1,     propertyB: 2,     .     . // more properties here     .     propertyM: 13 };  var objectB = {     propertyN: 14,     propertyO: 15,     .     . // more properties here     .     propertyZ: 26 };  var objectC = {...objectA, ...objectB}; // this is the answer  var a = objectC.propertyA;  var n = objectC.propertyN; 

Based on this article: https://blog.mariusschulz.com/2016/12/23/typescript-2-1-object-rest-and-spread

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In addition, the order of the variables in the decomposition matters. Consider the following:

var objectA = {     propertyA: 1,     propertyB: 2, // same property exists in objectB     propertyC: 3 };  var objectB = {     propertyX: 'a',     propertyB: 'b', // same property exists in objectA     propertyZ: 'c' };  // objectB will override existing properties, with the same name, // from the decomposition of objectA var objectC = {...objectA, ...objectB};   // result: 'b' console.log(objectC.propertyB);   // objectA will override existing properties, with the same name, // from the decomposition of objectB var objectD = {...objectB, ...objectA};   // result: '2' console.log(objectD.propertyB);