Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

253
Views
Sort an array of objects in typescript?

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!

about 4 years ago · Santiago Trujillo
3 answers
Answer question

0

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;
});
about 4 years ago · Santiago Trujillo Report

0

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));
about 4 years ago · Santiago Trujillo Report

0

    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;
    });
about 4 years ago · Santiago Trujillo Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!