Access Key From Dynamic Array Object in Javascript Without knowing key

Question

var data = [
    { 'Promotions 1': 2192, 'dates': '2021-02-04' },
    { 'Promotions 23': 767, 'dates': '2021-02-06' },
    { 'Promotions 12': 2264,'dates': '2021-02-08' },
];

The Issue is Promotions are dynamic key only access dates key-value. I want output like this below. I want that on unique every dates with all promotions Expected Output

var totalUserJoined = [
    { 
        'Promotions 1': 2192,
        'Promotions 23': 767,
        'Promotions 12': 2264 ,
        'dates': '2021-02-04'
    },
    { 
        'Promotions 1': 2192,
        'Promotions 23': 767,
        'Promotions 12': 2264 ,
        'dates': '2021-02-06'
    },
    { 
       'Promotions 1': 2192,
        'Promotions 23': 767,
        'Promotions 12': 2264 ,
        'dates': '2021-02-08'
    }
];

What did you try? What did go wrong?

Hansanho
– Hansanho

2021-03-04 08:47:59 +00:00
Commented Mar 4, 2021 at 8:47 — Hansanho
– Hansanho, Commented Mar 4, 2021 at 8:47
Object.keys(data) / Object.entries(data)

user5734311
– user5734311

2021-03-04 08:51:30 +00:00
Commented Mar 4, 2021 at 8:51 — user5734311
– user5734311, Commented Mar 4, 2021 at 8:51

Nick Parsons · Accepted Answer · 2021-03-04 08:55:44Z

2

You can use .map() and destructuring assignment to get all keys other than dates into their own object. Once you have an array of objects only containing the promotion keys (ie: the none dates), you can merge them all together using Object.assign() to form one object. Then you can use .map() on your original data array to merge the promotions object with every date:

const data = [ { 'Promotions 1': 2192, 'dates': '2021-02-04' }, { 'Promotions 23': 767, 'dates': '2021-02-06' }, { 'Promotions 12': 2264,'dates': '2021-02-08' }, ];

const promotions = Object.assign(...data.map(({dates, ...rest}) => rest));
const res = data.map(({dates}) => ({...promotions, dates}));

console.log(res);

answered Mar 4, 2021 at 8:55

Nick Parsons

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1 Comment

Shubham Seth Over a year ago

thanks stackoverflow.com/users/5648954/nick-parsons it works perfect

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