Question 246290 in Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript – map json object to key/value objects

I have a json as follows.

order = 
    {
        requests: [
            { code: 'T1', id: '123', requestStatus: 'request' },
            { code: 'T2', id: '321', requestStatus: 'request' }
        ]
    }

I want to get an object with key/value pair as follows.

ids['123']='T1'
ids['321']='T2'

I tried with this.

const order = 
{
    requests: [
        { code: 'T1', id: '123', requestStatus: 'request' },
        { code: 'T2', id: '321', requestStatus: 'request' }
    ]
};

let id = order['requests'].map((r) => [r.code] = r.id);

console.log(ids['123'] == 'T1');
console.log(ids['321'] == 'T2');

I am getting an error ‘object null is not iterable (cannot read property Symbol(Symbol.iterator))’.

Can someone show me how to do it?

Thanks.

Tags:

3

Answers


  1. 0 
    const order = {
  requests: [
    {code: 'T1', id: '123', requestStatus: 'request'},
    {code: 'T2', id: '321', requestStatus: 'request'}
  ]
};


let ids = {};
order['requests'].forEach(({code, id}) => ids[id] = code);

console.log(ids);

    prints

    { '123': 'T1', '321': 'T2' }

    Note that I use destructuring assignment to unpack objects in the lambda expression.

    Login or Signup to reply.
  2. 0

    You can use the Array.reduce method:

    order = {
  requests: [
    {code: 'T1', id: '123', requestStatus: 'request'},
    {code: 'T2', id: '321', requestStatus: 'request'}
  ]
}

const ids = order.requests.reduce((acc, req) => {
    // Add each property to the accumulator
    acc[req.id] = req.code;
    return acc;
}, {}); // Start with an empty object
    Login or Signup to reply.
  3. 0 
    const order = {
  requests: [
    { code: 'T1', id: '123', requestStatus: 'request' },
    { code: 'T2', id: '321', requestStatus: 'request' }
  ]
};

const ids = {};
order.requests.forEach((request) => {
  ids[request.id] = request.code;
});
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search