Answers for "array.sort js"

14

sorting array from highest to lowest javascript

// Sort an array of numbers 
let numbers = [5, 13, 1, 44, 32, 15, 500]

// Lowest to highest
let lowestToHighest = numbers.sort((a, b) => a - b);
//Output: [1,5,13,15,32,44,500]

//Highest to lowest
let highestToLowest = numbers.sort((a, b) => b-a);
//Output: [500,44,32,15,13,5,1]
Posted by: Guest on March-25-2020
5

sort array of objects javascript

list.sort((a, b) => (a.color > b.color) ? 1 : -1)
Posted by: Guest on July-23-2020
25

JS array sort

numArray.sort((a, b) => a - b); // For ascending sort
numArray.sort((a, b) => b - a); // For descending sort
Posted by: Guest on May-08-2021
13

javascript orderby

var items = [
  { name: 'Edward', value: 21 },
  { name: 'Sharpe', value: 37 },
  { name: 'And', value: 45 },
  { name: 'The', value: -12 },
  { name: 'Magnetic', value: 13 },
  { name: 'Zeros', value: 37 }
];

// sort by value
items.sort(function (a, b) {
  return a.value - b.value;
});

// sort by name
items.sort(function(a, b) {
  var nameA = a.name.toUpperCase(); // ignore upper and lowercase
  var nameB = b.name.toUpperCase(); // ignore upper and lowercase
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names must be equal
  return 0;
});
Posted by: Guest on May-30-2020
2

js index sorted

var test = ['b', 'c', 'd', 'a'];
var len = test.length;
var indices = new Array(len);
for (var i = 0; i < len; ++i) indices[i] = i;
indices.sort(function (a, b) { return test[a] < test[b] ? -1 : test[a] > test[b] ? 1 : 0; });
console.log(indices);
Posted by: Guest on March-08-2020
3

sort js

numArray.sort((a, b) => a - b); // For ascending sort
numArray.sort((a, b) => b - a); // For descending sort
Posted by: Guest on October-20-2020

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