Question

I have an array of objects that may or may not be populated with an array in it’s properties. Calling $.each(x,function(){}) on a null x results in a type error

try{
  var a = {};
  $.each(a.doesnotexist,function(k,v){})
} catch(e) {
  console.log(e.message)
}

I’ve seen suggestions of using $.extend on an empty object and passing that value to $.each(). Both of these seem to work for me and seem simpler, are there any downsides to either?

//some test values, not guaranteed properties are populated
var a = [
  {v: [1,2,3,4,5], w: [1,2,3,4,5,6]},
  {v: [1,2,3,4,5]},
  {w: [1,2,3,4,5]},
  {v: [1,2,3], w: []},
  {v: null}
];

for (i =0; i < a.length; i++) {if (a[i].v) $.each(a[i].v,function(k,v){});}
for (i =0; i < a.length; i++) {$.each(a[i].v||[],function(k,v){});}

The second seems cleanest to me.

Answer 1

You can make use of the in operator to check if a property exists in an object.

for (var i = 0; i < a.length; ++i) {
  if ("v" in a[i]){ 
    $.each(a[i].v, function(k,v) {
      ...
    });
  }
}

And, use instanceof operator to check against the correct variable type.

if ("v" in a[i] && a[i].v instanceof Array) { ... } 

In contrast to the short-circuit evaluation, this one not susceptible to truthy values.

$.each(a[i].v || [], ...);