Searched and tried and no luck so far.
var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ]
var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ]
What I want returned is:
var leftUsers = [{name: 'lauren', id: 35}, {name: 'dave', id: 28} ]
basically without the rich
object as this is a duplicate. I only care about the id
key.
I have tried:
newUsers.forEach((nUser) => {
likedUsers.forEach((lUser) => {
if (nUser.id !== lUser.id){
leftUsers.push(nUser)
}
})
})
but obviously this won't work as this will just add them all as soon as they don't match.
if possible would like an es6 solution using forEach/map/filter
thanks
With array.prototype.filter
to filter out items that exists in likedUsers
and array.prototype.findIndex
to check the existence, it should be:
var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ];
var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ];
var leftUsers = newUsers.filter(u => likedUsers.findIndex(lu => lu.id === u.id) === -1);
console.log(leftUsers);
You can do this with filter()
and some()
methods.
var newUsers = [{name: 'rich', id: 25}, {name: 'lauren', id: 35}, {name: 'dave', id: 28} ]
var likedUsers = [{name: 'derek', id: 39}, {name: 'rich', id: 25}, {name: 'brian', id: 38} ]
const result = newUsers.filter(e => !likedUsers.some(a => a.id == e.id));
console.log(result)
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