递归比较javascript中的一组对象

Question

I make an ajax call that returns an array of objects. The array can contain 1 object or 20, it is not known before making the call.

The objects will have the same properties but different values. I can compare two objects like this:

function compareObjects(x, y) {
   var objectsAreSame = true;
   for(var propertyName in x) {
      if(x[propertyName] !== y[propertyName]) {
         objectsAreSame = false;
         break;
      }
   }
   return objectsAreSame;
}

How can I extend this to compare several objects or is there a better way to compare ?

Edit#1 Similar question here: Object comparison in JavaScript

However, I am not trying to compare two objects, I am trying to find the shortest possible way to compare an unknown array of objects. I apologize for any confusion.

Edit #2 I am not asking how to compare two objects. I am asking how to take an array of unknown quantities and compare every object inside and check if every object is the same.

Like arrayofObjects[obj1, obj2...objn] simplest way to say obj1, through objn are exactly the same.

Answer 1

Keeping your fonction as is (it's very simplistic but let's assume it works in your use case):

const sameArray = (newArray.length === oldArray.length) && newArray.every((item, index) =>
  compareObjects(item, oldArray[index])
)

Answer 2

Stringify 'em!

var same = JSON.stringify(x) === JSON.stringify(y);

Edit: OP's slight edit can change this answer a bit. I thought he wanted to know if obj1 === obj2 (if they're exactly the same). This will do that well but not if properties can be in different orders and still be considered "same" for his use case.

Answer 3

I would do this with a generic method Object.prototype.compare() as follows

Object.prototype.compare = function(o){
  var ok = Object.keys(this);
  return typeof o === "object" && ok.length === Object.keys(o).length ? ok.every(k => this[k] === o[k]) : false;
};

Answer 4

You can use Array.prototype.every() and compare all objects in the array using the first object as reference for the comparison.

function compareArray(array) {
  return array.every(compareObjects.bind(null, array[0]));
}

 function compareObjects(x, y) { var objectsAreSame = true; for(var propertyName in x) { if(x[propertyName] !== y[propertyName]) { objectsAreSame = false; break; } } return objectsAreSame; } function compareArray(array) { return array.every(compareObjects.bind(null, array[0])); } var array1 = [ { a: 1 }, { a: 1 }, { a: 1 }, { a: 1 } ]; var array2 = [ { a: 1 }, { b: 1 }, { c: 1 }, { d: 1 } ]; console.log(compareArray(array1)); console.log(compareArray(array2));

Answer 5

In case if you need to compare array of objects and objects can contain arrays and each array can contain something different from object, you can use this:

const isEqual = (a, b) => {
    if (typeof a === "object" && typeof b === "object") {
        if (Array.isArray(a) && Array.isArray(b)) {
            return a.every((item, index) => isEqual(item, b[index]));
        }
        return Object.entries(a).toString() === Object.entries(b).toString();
    }

    return a === b;
};

For array this function will called recursively