Sort an array of objects in typescript? [duplicate]

Question

How do I sort an array of objects in TypeScript?

Specifically, sort the array objects on one specific attribute, in this case nome ("name") or cognome ("surname")?

/* Object Class*/
export class Test{
     nome:String;
     cognome:String;
}

/* Generic Component.ts*/
tests:Test[];
test1:Test;
test2:Test;

this.test1.nome='Andrea';
this.test2.nome='Marizo';
this.test1.cognome='Rossi';
this.test2.cognome='Verdi';

this.tests.push(this.test2);
this.tests.push(this.test1);

thx!

Answer 1

Simplest way for me is this:

Ascending:

arrayOfObjects.sort((a, b) => (a.propertyToSortBy < b.propertyToSortBy ? -1 : 1));

Descending:

arrayOfObjects.sort((a, b) => (a.propertyToSortBy > b.propertyToSortBy ? -1 : 1));

---

In your case, Ascending:

testsSortedByNome = tests.sort((a, b) => (a.nome < b.nome ? -1 : 1));
testsSortedByCognome = tests.sort((a, b) => (a.cognome < b.cognome ? -1 : 1));

Descending:

testsSortedByNome = tests.sort((a, b) => (a.nome > b.nome ? -1 : 1));
testsSortedByCognome = tests.sort((a, b) => (a.cognome > b.cognome ? -1 : 1));

Answer 2

It depends on what you want to sort. You have standard sort funtion for Arrays in JavaScript and you can write complex conditions dedicated for your objects. f.e

var sortedArray: Test[] = unsortedArray.sort((obj1, obj2) => {
    if (obj1.cognome > obj2.cognome) {
        return 1;
    }

    if (obj1.cognome < obj2.cognome) {
        return -1;
    }

    return 0;
});

Answer 3

    const sorted = unsortedArray.sort((t1, t2) => {
      const name1 = t1.name.toLowerCase();
      const name2 = t2.name.toLowerCase();
      if (name1 > name2) { return 1; }
      if (name1 < name2) { return -1; }
      return 0;
    });

Answer 4

You could use a function with a generic type:

const sortArrayOfObjects = <T>(
  data: T[],
  keyToSort: keyof T,
  direction: 'ascending' | 'descending' | 'none',
) => {
  if (direction === 'none') {
    return data
  }
  const compare = (objectA: T, objectB: T) => {
    const valueA = objectA[keyToSort]
    const valueB = objectB[keyToSort]

    if (valueA === valueB) {
      return 0
    }

    if (valueA > valueB) {
      return direction === 'ascending' ? 1 : -1
    } else {
      return direction === 'ascending' ? -1 : 1
    }
  }

  return data.slice().sort(compare)
}

Now if you use the function to sort array of objects, you can specify the object attribute you want to sort by.

const array = [
  { id: 2, name: "name2" },
  { id: 1, name: "name1" },
  { id: 3, name: "name3" }
];

const sortedArray = sortArrayOfObjects(array, "id", "ascending")

console.log(sortedArray)
//sorted ascending by id
// [
//   { id: 1, name: "name1" },
//   { id: 2, name: "name2" },
//   { id: 3, name: "name3" }
// ]

Using generic type and 'keyOf T' as type of attribute helps us to select only available object attributes, in this case, "id" and "name". typescript helper

CodeSandbox example

Answer 5

this.tests.sort(t1,t2)=>(t1:Test,t2:Test) => {
    if (t1.nome > t2.nome) {
        return 1;
    }

    if (t1.nome < t2.nome) {
        return -1;
    }

    return 0;
}

have you tried sth like this?

Answer 6

[{nome:'abc'}, {nome:'stu'}, {nome:'cde'}].sort(function(a, b) {
  if (a.nome < b.nome)
    return -1;
  if (a.nome > b.nome)
    return 1;
  return 0;
});

Answer 7

I often need to sort on a single field in a data set result, say Id and the shortest way of writing that is this

data.sort((a, b) => a.Id - b.Id)
    .forEach(row => {
           //something with row
     });

Answer 8

var arr = [110,5,72,14,12]

Ascending Order Sort

arr.sort(( a, b ) => a > b ? 1 : -1 )

Descending Order Sort

arr.sort(( a, b ) => a < b ? 1 : -1 )

Answer 9

you can use this method.

let sortedArray: Array<ModelItem>;
sortedArray = unsortedArray.slice(0);
sortedArray.sort((left, right) => {
    if (left.id < right.id) return -1;
    if (left.id > right.id) return 1;
    return 0;
})

Answer 10

I am using angular 7 and I tried pipe but didn't work and I tried this and got the output right.

let array of objects be in object result

results:any ={
 "providers": [
  {
    "name": "AAA",
    "error": {
      "success": true,
      "message": "completed"
    },
    "quote_id": "BA503452VPC0012790",
    "quotes": {
      "comprehensive": {
        "premiumBreakup": {
          "premium": "17398.0",
        }
      },
    }
  },
  {
    "name": "Fi",

    "error": {
      "success": true,
      "message": "completed"
    },
    "quotes": {
      "comprehensive": {
        "premiumBreakup": {
          "premium": "27957.00"              
        }
      },
    }
  },
]
}

On clicking sort function

<button class="t-sort-optn" (click)="sortByPremium()">Premium</button>

To sort based on the premium value

sortByPremium(){
    var items = this.results.providers;
    console.log("Array",items);
    items.sort(function (a, b) {
    return a.quotes.comprehensive.premiumBreakup.premium - b.quotes.comprehensive.premiumBreakup.premium;
    });
    console.log("Array Sorted",items)
}

Answer 11

This is how it worked for me

.sort((a, b) => {
  const a2 = a as unknown as Type;
  const b2 = b as unknown as Type;
  if (a2.something > b2.something) {
    return 1;
  }
  
  if (a2.something < b2.something) {
    return -1;
  } 

  return 0;
})

Answer 12

To make it more both readable and reusable, we have created a helper function:

export const by = <T>(attribute: keyof T) => {
  return (one: T, two: T) => {
    if (one[attribute] < two[attribute]) {
      return -1;
    } else if (one[attribute] > two[attribute]) {
      return 1;
    } else {
      return 0;
    }
  };
};

Then we can use it like this:

const users = [ /* items here */ ];
const sorted = users.sort(by('name'));

Answer 13

consider your array as myArray,

myArray.sort(( a, b ) => a > b ? 1 : -1 )