How to use input value as array index

Question

I'm actaully out of ideas. I've been reading around about this problem but I didn't find any answer that suits my need. So here is the scenario

I have some inputs where I get a value by their id

<input type="hidden" id="name" value="Taco house" />
<input type="hidden" id="pin" value="Mexican" />

And I try to use the value as index to build an array

var name = $('#name').val();
var pin = $('#pin').val();

markersData = {
    pin: [ //here it should put 'Mexican' instead it puts 'pin'
    {
        name: name,
        location_latitude: 43.5391318, 
        location_longitude: 10.3020171,
        //other nodes here
    }
    ]
};

If I print pin variable before passing to the array it prints its actual value, but once I pass to the array the value is lost and the index became pin.

For sure this is an obvious question but since I'm not that good in jquery I need some help.

Your problem is not at all clear, but have you tried var foo = markersData.pin[parseInt(pin, 10)] — Rory McCrossan
– Rory McCrossan, Commented Oct 3, 2016 at 13:40
Could you post some more info? hard to understand the problem from your description — Aidan Gee
– Aidan Gee, Commented Oct 3, 2016 at 13:42
You do not have a pin with attribute pin. You have a named pin. So what is name:name looking like? name:"Taco house" ? — mplungjan
– mplungjan, Commented Oct 3, 2016 at 13:44

vincenth · Accepted Answer · 2016-10-03 13:53:18Z

2

This should do the trick :

var name = $('#name').val();
var pin = $('#pin').val();
markersData = {};
markersData[pin] = [{
    name: name,
    location_latitude: 43.5391318,
    location_longitude: 10.3020171,
    //other nodes here
}];

Your way won't work, because the parser consider pin as a string, not as the content of your "pin" variable. Neither can you do markersData.pin for the same reason.

As far as I know (I might be wrong), the only way to do this is to use square brackets, which might be a bit confusing since it looks like adding an index to an array. Hope this helps

edited Oct 3, 2016 at 13:53

answered Oct 3, 2016 at 13:48

vincenth

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17 Comments

Zack Over a year ago

Is markersData[pin] the same as markersData.pin? Please explain what you are actually doing here to help the asker understand why this should "do the trick".

vincenth Over a year ago

@Zack, I was actually updating my answer to add some clarification. It should also answer your question.

Fabio Over a year ago

This is actually working but I don't understand why I cannot use variable to build an array index in my previous way though

vincenth Over a year ago

@Fabio Someone can provide you with a better explanation, sinceI am not versed in javascript inner working, but from what I understand, the parser does not see pin as a variable when you use it as an object keys. I think it is intended to prevent confusion, JS can't know if you want to use a string or a variable you declared.

Fabio Over a year ago

I see, i'm not expert too, i know it should work properly in php

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