Check if any object in the object array contains all key / value pairs in another object

Question

I am trying to write a function that looks through an array of objects (the first argument) and returns an array of all objects that contain all the key / value pairs of the given object (second argument).

My code below only works if the object source(second argument) contains one key / value pair. If the object sourcehas two or more key / value pairs, the result is not as expected.

How to account for multiple key / value pairs in an object source?

function findObjects(collection, source) {
  var result = [];

  for (i=0; i<collection.length; i++) {
    for (var prop in source) {
      if (collection[i].hasOwnProperty(prop) && collection[i][prop] == source[prop]) {
        console.log('Collection\ object ' + [i] + ' contains the source\ key:value pair ' + prop + ': ' + source[prop] + '!');
        result.push(collection[i]);
      } else {
        console.log('fail');
      }
    }
  }

  console.log('The resulting array is: ' + result);
  return result;
}

findObjects([{ "a": 1, "b": 2 }, { "a": 1 }, { "b": 2, "c": 2 }], { "a": 1, "b": 2 });

// only the first object should be returned since it contains both "a":1 and "b":2

Answer 1

function findObjects(collection, source) {
    var result = [];

    for (i=0; i<collection.length; i++) {
        var matches = true;
        for (var prop in source) {
            if (!collection[i].hasOwnProperty(prop) || collection[i][prop] !== source[prop]) {
                matches = false;
                break;
            }
        }
        if (matches) {
            result.push(collection[i]);
        }
    }
    return result;
}

, , , result.

Answer 2

, Array#map

map() .

Array#every

every() , , .

Object.keys.

Object.keys() , for...in ( , for-in ).

function findObjects(collection, source) {
    var keys = Object.keys(source);          // get all properties of source
    return collection.filter(function (c) {  // filter collection
        return keys.every(function (k) {     // check every key
            return c[k] === source[k];       // compare value of collection and source
        });
    });
}

console.log(findObjects([{ "a": 1, "b": 2 }, { "a": 1 }, { "b": 2, "c": 2 }], { "a": 1, "b": 2 }));

Hide result

ES6 (: )

function (x) { return y; }

          x =>        y

function findObjects(collection, source) {
    const keys = Object.keys(source);
    return collection.filter(c => keys.every(k => c[k] === source[k]));
}

console.log(findObjects([{ "a": 1, "b": 2 }, { "a": 1 }, { "b": 2, "c": 2 }], { "a": 1, "b": 2 }));

Hide result

Answer 3

The solution uses a function Object.keys, Array.forEachand Array.some:

function findObjects(arr, source) {
    var keys = Object.keys(source), result = [];
    arr.forEach(function(v){
        !keys.some((k) => source[k] !== v[k]) && result.push(v);
    });

    return result;
}

console.log(findObjects([{ "a": 1, "b": 2 }, { "a": 1 }, { "b": 2, "c": 2 }], { "a": 1, "b": 2 }));
// it will output [{"a":1,"b":2}]