jQuery find element in array

Question

I am trying to find an element in the array. Here is the fiddle:

var selectedVals = [630, 85];
var registeredVals = [17, 630, 85];
var newList = [];

$(selectedVals).each(function () {
    //registeredVals.splice($.inArray(this, registeredVals), 1);
    var num = this;
    alert("Value="+ num + " Array=" + registeredVals);
    //alert(registeredVals.indexOf(num));
    alert($.inArray(this, registeredVals));

    if($.inArray(this, registeredVals) == -1)
         newList.push(this);
    })

https://jsfiddle.net/programmedprojects/7m9zrjwL/

I tried .indexOf and $.inArray to find the element in loop.

Shouldn't the index be greater than -1 as the element is in the array?

Basically, i am trying to find the duplicates. I just opted this way as i didn't find the desired behavior. There are two multiselect list boxes, on a button click, i want to move the items from one to another without removing from the parent and not add duplicates to the second list — Programmer
– Programmer, Commented Apr 6, 2018 at 17:40

Eddie · Accepted Answer · 2018-04-06 17:52:15Z

1

If you want to know what elements present on both selectedVals and registeredVals, you can use filter() and use includes() to check if a variable exists on an array.

var selectedVals = [630, 85, 99];
var registeredVals = [17, 630, 85];

var newList = selectedVals.filter(o => registeredVals.includes(o));

console.log(newList);

Without ES6

var selectedVals = [630, 85, 99];
var registeredVals = [17, 630, 85];
var newList = [];

for (var i = 0; i < selectedVals.length; i++) {
  if (registeredVals.indexOf(selectedVals[i]) !== -1) newList.push(selectedVals[i]);
}

console.log(newList);

edited Apr 6, 2018 at 17:52

answered Apr 6, 2018 at 17:44

Eddie

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3 Comments

Programmer Over a year ago

I tried this but looks like IE doesn't support => and i get SCRIPT1002: Syntax error. Even the above doesn't run on IE 11.

Programmer Over a year ago

Thanks for the solution. But what was wrong with $.each loop? Why did the index returned wrong?

Eddie Over a year ago

I think the whole use of jQuery is wrong.You should use $.each instead of $(selectedVals)

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