jquery create object dynamic not working

Question

I need the output as

{       'US-CA': '#084365',
        'US-TX': '#084365',
        'US-CO': '#00a2e8',
        'US-NM': '#00a2e8',
        'US-WY': '#00a2e8',
        'US-NE': '#00a2e8'
    }

For this I used the following code:

   var output = [];

$('.vectordata').find('.varc').each(function(i){
            var t = $(this);
            regioncode = t.find('.regioncode').val();
            color = t.find('.color').val();
                    var obj2 = {}
                    obj2[regioncode] = color;
                    output.push(obj2);

    }

But the output I received is

Please help me fix the dynamic object creation

You are getting array of objects in output but you need literal object. This is because you are pushing single object in one array — KARAN LAGALWAR
– KARAN LAGALWAR, Commented Jul 18, 2018 at 5:53

Amit Wagner · Accepted Answer · 2018-07-18 05:50:49Z

2

you dont need the array just put it in the object

 var output = {};

$('.vectordata').find('.varc').each(function(i){
            var t = $(this);
            regioncode = t.find('.regioncode').val();
            color = t.find('.color').val();
            output[regioncode] = color;
                   

    }

answered Jul 18, 2018 at 5:50

Amit Wagner

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Comments

Rashedul.Rubel · Accepted Answer · 2018-07-18 05:58:17Z

2

You want to get output as JSON object like below:

{       'US-CA': '#084365',
        'US-TX': '#084365',
        'US-CO': '#00a2e8',
        'US-NM': '#00a2e8',
        'US-WY': '#00a2e8',
        'US-NE': '#00a2e8'
}

so, in order to serve your purpose you need to declare an object and put the values into object as {key:value} pair. Don't need any array to store the object again. You can follow samuellawrentz answer.

answered Jul 18, 2018 at 5:58

Rashedul.Rubel

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Comments

samuellawrentz · Accepted Answer · 2018-07-18 05:50:04Z

1

       var output = {};

    $('.vectordata').find('.varc').each(function(i){
                var t = $(this);
                regioncode = t.find('.regioncode').val();
                color = t.find('.color').val();
                        output[regioncode] = color;

        }

console.log(output);

You were pushing objects into Array. Make output as object.

answered Jul 18, 2018 at 5:50

samuellawrentz

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