Store different numbers into an array

Question

Lets say you have a situation where you get 3 numbers from different sources, I wanna merge them together in 1 array and came up with this. Its almost do the trick but I can't combine the numbers into 1 array

$(n + " " + 'ul').each(function(){
 var arr = ( $(this).children('li').size());
 var test = (""+arr).split("");
 console.log(test)
});
//Outcome in console
"1"
"2" 
"3"
//Outcome in console what I need
"1, 2, 3"

Something about this line isn't right, I know the split doesn't work but I can't come up with something better to achieve the result.

var test = (""+arr).split("");

• Barmar: .size() doesn't return an array, it returns a number. I don't understand what you're doing with that.
• Dennis: Yes I know but it takes the UL and takes the amount of list items from 1 single UL

Answer 1

If you want to have the result as one single string (and not an array), I would use map and then join the result:

var n = "div";
var res = $(n + " " + 'ul').map(function(){
 return $(this).children('li').length; // size() is deprecated
}).get().join(", ");
alert(res); // "1, 2, 3"

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="one">
 <ul>
 <li></li>
 </ul>
</div>
<div class="two">
 <ul>
 <li></li>
 <li></li>
 </ul>
</div>
<div class="three">
 <ul>
 <li></li>
 <li></li>
 <li></li>
 </ul>
</div>

Answer 2

Should it be something like this? You are creating a new array every loop through there:

var test = [];
$(n + " " + 'ul').each(function(){
 test.push($(this).children('li').size()); 
});
console.log(test)

Answer 3

Use:

 $(n + " " + 'ul').each(function(){
 var arr = ( $(this).children('li').size());
 var test = arr.toString();
 });
 console.log(test);

Answer 4

You should look at Array join

var arr = [];
$(n + " " + 'ul').each(function(){
 arr.push($(this).children('li').size()));
});
var test = arr.join(", ");
console.log(test);

• Barmar: arr is not an array, you can't use .join on it.
• Mritunjay: @Barmar updated my answer. I thought of just idea last time.