How does one return the object values only from each object of an array of objects?

Question

So I am running a NodeJS Serverless instance, using Knex.js as my DB middleware. When attempting to return JUST the value from a request to Knex, I always get [ { vrn: 'xx12xyz' } ]. I have tried using Object(), does anyone have a quick one liner or small helper function that can help me flatten this into a direct array such that I can pick out JUST the value?

This is the code I am using:

const vrnList = await db('vehicles').select('vrn').where('id', '=', '1');
    console.log("The quick car with the VRN ", Object.values(vrnList), " overtook me on the freeway");

TL;DR: Splitting the object and array id from the values

Luís Assunção · Accepted Answer · 2021-10-10 23:03:58Z

1

Try this one:

let onlyValues = vrnList.map(result => result.vrn);

answered Oct 10, 2021 at 23:03

Luís Assunção

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Comments

Peter Seliger · Accepted Answer · 2021-10-10 23:25:27Z

1

Since the OP seems to retrieve the value(s) key agnostic via Object.values which returns an array of values, the OP in addition needs to utilize flatMap ...

vrnList.flatMap(item => Object.values(item))

console.log(
  [ { vrn: 'xx12xyz' } ].flatMap(item => Object.values(item))
);
console.log(
  [
    { vrn: 'xx12xyz' },
    { vrn: 'xx98abc' },
    { vrn: 'yy34xyz' },
  ].flatMap(item => Object.values(item))
);

.as-console-wrapper { min-height: 100%!important; top: 0; }

edited Oct 10, 2021 at 23:25

answered Oct 10, 2021 at 23:07

Peter Seliger

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4 Comments

JPA Over a year ago

All fixed - Thank you. I was left with an array but using the toString() function fixed it. Thanks!

Peter Seliger Over a year ago

@JPA ... maybe the OP considers giving an own answer of how the problem got solved. I personally doubt that a toString on the provided data structure actually was the only thing needed in order to fix all issues.

JPA Over a year ago

No what I meant was that your answer, partnered with toString(), made it work

Peter Seliger Over a year ago

@JPA ... "No what I meant was that your answer, partnered with toString(), made it work" ... good to hear. And at SO it is considered to be a nice gesture from the one who got help, to accept the answer which was the most helpful in solving the OP's problem.

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