Filter an array containing objects based on another array containing objects in angular js

Question

I am working on angular js and i am new to it. So I am stuck at a problem where I have to subtract two arrays containing objects.

For Eg.

var all = [{id:'1',name:'A'},{id:'2',name:'B'},{id:'3',name:'C'},{id:'4',name:'D'}];
var old = [{id:'1',name:'A',state:'healthy'},{id:'3',name:'C',state:'healthy'}];

var newArray = [];

Now, I wanted to populate the 'newArray' variable with the objects that are not existing in 'old' list as below

newArray = [{id:'2',name:'B'},{id:'4',name:'D'}]

Is there anyway to achieve this in angular js? Thanks

Answer 1

try this

 var all = [{ id: '1', name: 'A' }, { id: '2', name: 'B' }, { id: '3', name: 'C' }, { id: '4', name: 'D' }]; var old = [{ id: '1', name: 'A', state: 'healthy' }, { id: '3', name: 'C', state: 'healthy' }]; var newlist = all.filter(function(a) { return old.filter(function(o) { return o.id == a.id }).length == 0 }) document.write('<pre>' + JSON.stringify(newlist, 0, 4) + '</pre>')

Answer 2

You can do it using Array.prototype.forEach() and Array.prototype.some()

 var all = [{id:'1',name:'A'},{id:'2',name:'B'},{id:'3',name:'C'},{id:'4',name:'D'}]; var old = [{id:'1',name:'A',state:'healthy'},{id:'3',name:'C',state:'healthy'}]; var newArr = []; all.forEach(function(e) { if(!old.some(s => s.id == e.id)) { newArr.push(e); } }); document.write('<pre>' + JSON.stringify(newArr, 0, 2) + '</pre>');

Answer 3

A proposal with linear complexity.

 var all = [{ id: '1', name: 'A' }, { id: '2', name: 'B' }, { id: '3', name: 'C' }, { id: '4', name: 'D' }], old = [{ id: '1', name: 'A', state: 'healthy' }, { id: '3', name: 'C', state: 'healthy' }], result = function (array1, array2) { var o = {}; array2.forEach(function (a) { o[a.id] = true; }); return array1.filter(function (a) { return !o[a.id]; }); }(all, old); document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

Answer 4

This is not at all related to AngularJS . This is a javascript problem, and you can easily achieve this result by using the following code-

var new = all.filter(function(element) { 
    var res = $.grep(old, function(el) { 
        return el.id == element.id; 
    }); 
    if(res.length == 0) return element; 
});