Issue

This Content is from Stack Overflow. Question asked by asfafafags

I’m trying to sort the cars by model and month and it works well.
But I have an average month 00 which I want to be the last record of each model instead of the first.

I have no idea how to achieve this. I have created a snippet. Any help please?

let cars = [{
    model: 'Ford',
    month: '12'
  }, {
    model: 'Ford',
    month: '03'
  }, {
    model: 'Ford',
    month: '01'
  }, {
    model: 'Ford',
    month: '00'
  },
  {
    model: 'Ford',
    month: '05'
  },
  {
    model: 'Fiat',
    month: '05'
  },
  {
    model: 'Fiat',
    month: '01'
  },
  {
    model: 'Fiat',
    month: '00'
  }
];

console.log(`Unsorted cars: ${JSON.stringify(cars)}`);

const sorted = cars.sort((a, b) => a.model.localeCompare(b.model) || a.month.localeCompare(b.month));

console.log(`Sorted cars: ${JSON.stringify(sorted)}`);

Expected:

[{“model”:”Fiat”,”month”:”01″},{“model”:”Fiat”,”month”:”05″}, {“model”:”Fiat”,”month”:”00″},{“model”:”Ford”,”month”:”01″},{“model”:”Ford”,”month”:”03″},{“model”:”Ford”,”month”:”05″},{“model”:”Ford”,”month”:”12″}, {“model”:”Ford”,”month”:”00″}]

Solution

You could special-case the 00 month in your sort comparator function:

let cars = [
  { model: 'Ford', month: '12' },
  { model: 'Ford', month: '03' }, 
  { model: 'Ford', month: '01' },
  { model: 'Ford', month: '00' },
  { model: 'Ford', month: '05' },
  { model: 'Fiat', month: '05' },
  { model: 'Fiat', month: '01' },
  { model: 'Fiat', month: '00' }
];

console.log(`Unsorted cars: ${JSON.stringify(cars)}`);

const sorted = cars.sort((a, b) =>
  a.model.localeCompare(b.model) ||
  (a.month == '00' ? 1 : (b.month == '00' ? -1 : a.month.localeCompare(b.month)))
);

console.log(`Sorted cars: ${JSON.stringify(sorted)}`);

// alternate solution without nested ternary operator
const sorted2 = cars.sort((a, b) =>
  a.model.localeCompare(b.model) ||
  a.month == '00' ||
  (b.month == '00' ? -1 : a.month.localeCompare(b.month))
);

console.log(`Sorted cars: ${JSON.stringify(sorted2)}`);

This Question was asked in StackOverflow by asfafafags and Answered by Nick It is licensed under the terms of CC BY-SA 2.5. - CC BY-SA 3.0. - CC BY-SA 4.0.