Why does If statement throw an undefined error when I am checking if its undefined?

Question

Why does my else if statement cause the parser to throw an undefined error? I know that it is undefined which is why I am checking, I want it to hit the else block in this case, because this function is used in mutiple places the data that is being passed in is not always the same. So I am trying to check which type of data is being passed in.

if(icon){
    dv.find(".mobCol>.image").html(generateIcon(icon));
}
else if(location[0].properties !== undefined){
    dv.find(".mobCol>.image").html(generateIcon(location[0].properties.image));
    data = location[0];
}
else{
    dv.find(".mobCol>.image").html(generateIcon(location.features[0].properties.image));
    data = location.features[0];
}

Because you're checking at least one level past where the `undefined` is. — Dave Newton, May 07 '18 at 15:20

Martijn · Accepted Answer · 2018-05-07T18:09:31.280

4

Because location[0] is undefined. It's a bit odd, but your check should be:

if( location[0] && location[0].properties){ /* .. */}

I've omitted the ===undefined, thats not needed. It tests if it is something is thruthy or falsey This is a common way to check variables in JS.

edited May 07 '18 at 18:09

answered May 07 '18 at 15:19

Martijn

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This worked for me! Thank you! – Caitlyn Daly May 07 '18 at 15:23
"If something is set, it'll return true" is false though. If location[0] is null, false, 0, NaN or empty string it will also return false. – FINDarkside May 07 '18 at 15:28
Rephrazed it with a link :) – Martijn May 07 '18 at 18:09

Why does If statement throw an undefined error when I am checking if its undefined?

1 Answers1