CODEWITHSUNDEEP
javascriptjquery
Mathematics
Mathematics

Reputation: 7628

Getting Name initials using JS

I would like to extract initials from a string, like:

Name = FirstName LastName 
Initials =  FL

I can get the above result using this,

const initials = item
    .FirstName
    .charAt(0)
    .toUpperCase() +
  
    item
    .LastName
    .charAt(0)
    .toUpperCase();

But now my requirements are changed as if name only consist of 1 word or more then 2, so in following cases how can I get initials as per my requirements,

FullName =  FU
FirstName MiddleName LastName = FL
1stName 2ndName 3rdName 4thName 5thName = 15

How can I get above initials from a string in JS?

Also now I only have item.Name string as an input

Upvotes: 94

Views: 147569

Answers (30)

NMI
NMI

Reputation: 538

const initial = username.match(/\b(\w)/g).join('')

Upvotes: 9

Cuminato
Cuminato

Reputation: 542

There are many complicated answers here...

Simpler solution with Clean code!

I know this question has been around for a while, but I believe this solution is good enough to share with you :)

My goal with this code is to create something that is both intelligent and easy to read.

const names = [
  'Crystals', // -> CR
  'undisclosed desires', // -> UD
  'Feel so Close - Radio Edit', // -> FE
  ' ', // -> empty
  'Faint ', // -> FA .: Note the space after the name
]

function getInitials(fullName) {
  const [firstName, ...restNames] = fullName.toUpperCase().trim().split(' ')

  if (!restNames.length) {
    return firstName.substring(0,2)
  }

  const firstNameInitial = firstName[0]
  const lastNameInitial = restNames.pop()[0]

  return `${firstNameInitial}${lastNameInitial}`
}

console.log(names.map(getInitials))

The first line, transforms the string to uppercase, remove unwanted spaces (at the begginning and end) and split the name creating an array. Using the destructuring we recover the firstName and place the rest into a constant called restNames

const [firstName, ...restNames] = fullName.toUpperCase().trim().split(' ')

after that, we check if there's other names besides the first name, if not, we return the first two letters from it ending the function execution

if (!restNames.length) {
  return firstName[0].substring(0,2)
}

If we found out that restNames contains other names, we get the first name and the last name initials

const firstNameInitial = firstName[0]
const lastNameInitial = restNames.pop()[0]

Finally, we return the initials!

return `${firstNameInitial}${lastNameInitial}`

This code aims to be clear and efficient in providing the desired result

Upvotes: 1

Martinez
Martinez

Reputation: 458

const name = 'First Second Third'

name
    .split(' ')
    .slice(0, 2) // <= optional if only two letters are desired
    .map((name) => name[0])
    .join('')

// FS

Upvotes: 1

Mityu
Mityu

Reputation: 163

I am sorry if I misunderstand something, but most answers are so.... complicated.

My solution is (for international, actually hungarian names):

var regex = /[A-ZÍÉÁŰŐÚÖÜÓ]{1}/g
var monogram = 'Üveges Tóth Ödön'.match(regex).join('')

Upvotes: 0

Muhammad Abdullah
Muhammad Abdullah

Reputation: 483

A better way.

nameToInitials(name: string): string {
    const portions = name.split(' ')
        .map(val => val[0]);

    return portions.slice(0, 2)
        .reduce((a, b) => a + b, '').toUpperCase();
}

Upvotes: 0

Om Sutar
Om Sutar

Reputation: 1

var getInitials = function (string) {
var names = string.split(' '),
initials = names[0].substring(0, 1).toUpperCase()+'.';

if (names.length > 1) {
initials += names[names.length - 2].substring(0, 1).toUpperCase()+'.';
}
return initials=initials+names[names.length - 1].toUpperCase();
}

console.log(getInitials('Rama Krishna Narayan'));

var getInitials = function (string) {
    var names = string.split(' '),
    initials = names[0].substring(0, 1).toUpperCase()+'.';
    
    if (names.length > 1) {
    initials += names[names.length - 2].substring(0, 1).toUpperCase()+'.';
    }
    return initials=initials+names[names.length - 1].toUpperCase();
    }
 console.log(getInitials('Rama Krishna Narayan'));

Upvotes: 0

chickens
chickens

Reputation: 22314

Get First and Last Initial: John Doe Smith => JS

name.match(/(\b\S)?/g).join("").match(/(^\S|\S$)?/g).join("").toUpperCase()

Get All Initials: "John Doe Smith" => "JDS"

name.match(/(\b\S)?/g).join("").toUpperCase()

Get First and Last except get First 2 in case there is only first. (OP's question)

John => JO and "John Doe Smith" => "JS"

name.match(/(^\S\S?|\b\S)?/g).join("").match(/(^\S|\S$)?/g).join("").toUpperCase()

International Version: "Störfried Würgekloß" => "SW"

name.match(/(^\S\S?|\s\S)?/g).map(v=>v.trim()).join("").match(/(^\S|\S$)?/g).join("").toLocaleUpperCase()

Note: If the name contains , or other non word characters, you might use /w instead of /S or sanitize it beforehand

Upvotes: 57

test30
test30

Reputation: 3654

+ efficient
+ no loops
+ simplified branching (ternary operator only)
+ handles no-space cases (prints 2 chars)
+ no array memory allocation (actually no array processing) - requires trimmed string input

function getInitials(name) {
    const hasTokens = name.indexOf(' ') !== -1
    return name.substring(0, hasTokens ? 1 : 2) + (hasTokens ? name.charAt(name.lastIndexOf(' ') + 1) : '')
}

console.log(getInitials("A B"), 'AB')
console.log(getInitials("Abc Def"), 'AD')
console.log(getInitials("Abc Xyz"), 'AX')
console.log(getInitials("S Xyz"), 'SX')
console.log(getInitials("SXyz "), 'SX')
console.log(getInitials("T30"), 'T3')

Upvotes: 6

thorn0
thorn0

Reputation: 10427

function getInitials(name) {
  return (
    name
      .match(/(?<=\s|^)\p{L}\p{Mn}*/gu)
      ?.filter((el, i, array) => i === 0 || i === array.length - 1)
      .join("") || ""
  );
}

console.log(getInitials('ÇFoo Bar 1Name too ÉLong'));
console.log(getInitials('Q̈lice Hwerty')); // Q is followed by U+0308 (Combining Diaeresis)
console.log(getInitials('A Foo'));
console.log(getInitials('Bob'));

Safari doesn't yet support lookbehinds in regexes (see caniuse), so if Safari support is needed, it can be rewritten this way:

function getInitials(name) {
  return (
    name
      .match(/(\s|^)\p{L}\p{Mn}*/gu)
      ?.filter((el, i, array) => i === 0 || i === array.length - 1)
      .map(el => el.trimStart())
      .join("") || ""
  );
}

Upvotes: 1

Andross
Andross

Reputation: 246

Similar but slightly neater version of @njmwas answer:

let name = 'Fred Smith';
let initials = name.split(' ').reduce((acc, subname) =>
    acc + subname[0], '')
console.log(initials) // FS

Or, to include the abbreviation dots:

let name = 'Fred Smith';
let initials = name.split(' ').reduce((acc, subname) =>
    acc + subname[0] + '.', '')
console.log(initials) // F.S.

Upvotes: 3

mehparra
mehparra

Reputation: 446

Something more functional: D

  const getInitials = (string) => {
        const [firstname, lastname] = string.toUpperCase().split(' ');
        const initials = firstname.substring(0, 1);
        return lastname
          ? initials.concat(lastname.substring(0, 1))
          : initials.concat(firstname.substring(1, 2));
      };

console.log(getInitials('FirstName LastName')); // FL
console.log(getInitials('FirstName MiddleName LastName')); // FM
console.log(getInitials('FirstName')); // FI

Upvotes: 0

siwalikm
siwalikm

Reputation: 1875

There are some other answers which solve your query but are slightly complicated. Here's a more readable solution which covers most edge cases.

As your full name can have any number of words(middle names) in it, our best bet is to spit it into an array and get the initial characters from the first and last words in that array and return the letters together.

Also if your 'fullName' contains only one word, word at array[0] and array[array.length - 1] would be the same word, so we are handling that if the first if.

function nameToInitials(fullName) {
  const namesArray = fullName.trim().split(' ');
  if (namesArray.length === 1) return `${namesArray[0].charAt(0)}`;
  else return `${namesArray[0].charAt(0)}${namesArray[namesArray.length - 1].charAt(0)}`;
}

Sample outputs :

> nameToInitials('Prince') // "P"

> nameToInitials('FirstName LastName') // "FL"

> nameToInitials('1stName 2ndName 3rdName 4thName 5thName') // "15"

Upvotes: 8

ABHIJEET KHIRE
ABHIJEET KHIRE

Reputation: 2491

THIS IS THE SIMPLE UTILITY METHOD THAT HELPS TO GET THE INITIALS OF THE NAME BY SIMPLY PASSING THE NAME TO getInitials function // eg getInitials("harry potter") ==> "HP"

const getInitials = (name) => {
  var parts = name.split(' ')
  var initials = ''
  for (var i = 0; i < parts.length; i++) {
    if (parts[i].length > 0 && parts[i] !== '') {
      initials += parts[i][0]
    }
  }
  return initials.toUpperCase();
}

Upvotes: 0

Vandesh
Vandesh

Reputation: 6894

Common Avatar Use-case

Just surprised that none of the answers put Array.reduce() to good use.

const getInitials = (fullName) => {
  const allNames = fullName.trim().split(' ');
  const initials = allNames.reduce((acc, curr, index) => {
    if(index === 0 || index === allNames.length - 1){
      acc = `${acc}${curr.charAt(0).toUpperCase()}`;
    }
    return acc;
  }, '');
  return initials;
}

Run the snippet below to check the initials for different use cases -

const testNames = [
  'Albus Percival Wulfric Brian dumbledore', // AD
  'Harry Potter',  // HP
  'Ron', // R
  '', // <empty>
  'Çigkofte With Érnie', // ÇÉ
  'Hermione ', // H  (Notice that there is a space after the name) 
  'Neville LongBottom ' // NL (space after name is trimmed)
]

const getInitials = (fullName) => {
  const allNames = fullName.trim().split(' ');
  const initials = allNames.reduce((acc, curr, index) => {
    if(index === 0 || index === allNames.length - 1){
      acc = `${acc}${curr.charAt(0).toUpperCase()}`;
    }
    return acc;
  }, '');
  return initials;
}


console.log(testNames.map(getInitials));

Note

This one is for a widely used case for displaying names in Avatars, where you wouldn't want first name initial to be repeated twice and want to restrict the initials to a max of 2 letters

Upvotes: 27

Eugenia Bonocore
Eugenia Bonocore

Reputation: 31

I needed this today to act as method in my React code. I was getting the user name from the state as props. After that I just passed my method inside my component's props.

getUserInitials() {
  const fullName = this.props.user.name.split(' ');
  const initials = fullName.shift().charAt(0) + fullName.pop().charAt(0);
  return initials.toUpperCase();
 }

Upvotes: 2

Yuvraj Chauhan
Yuvraj Chauhan

Reputation: 239

You can use below one line logic:

"FirstName MiddleName LastName".split(" ").map((n,i,a)=> i === 0 || i+1 === a.length ? n[0] : null).join("");

Upvotes: 10

Shanoor
Shanoor

Reputation: 13682

Why no love for regex?

Updated to support unicode characters and use ES6 features

let name = 'ÇFoo Bar 1Name too ÉLong';
let rgx = new RegExp(/(\p{L}{1})\p{L}+/, 'gu');

let initials = [...name.matchAll(rgx)] || [];

initials = (
  (initials.shift()?.[1] || '') + (initials.pop()?.[1] || '')
).toUpperCase();

console.log(initials);

Upvotes: 95

knoefel
knoefel

Reputation: 6949

Easy way using ES6 Destructering:

const getInitials = string =>
  string
    .split(' ')
    .map(([firstLetter]) => firstLetter)
    .filter((_, index, array) => index === 0 || index === array.length - 1)
    .join('')
    .toUpperCase();

Upvotes: 0

Aregbesola OJ
Aregbesola OJ

Reputation: 11

To get the first name and last name initials, try using the function below.

const getInitials = string => {
    const names = string.split(' ');
    const initials = names.map(name => name.charAt(0).toUpperCase())
    if (initials.length > 1) {
        return `${initials[0]}${initials[initials.length - 1]}`;
    } else {
        return initials[0];
    }
};
console.log(getInitials("1stName 2ndName 3rdName 4thName 5thName")); // 15
console.log(getInitials("FirstName MiddleName LastName")); // FL

WHAT HAPPENED: The function splits the incoming string, ignores any name between the first & last names and returns their initials. In the case a single name is entered, a single initial is returned. I hope this helps, cheers.

Upvotes: 1

Thram
Thram

Reputation: 532

I saw a bunch of overcomplicated ways to do this. I'm really more into simplifying things as much as possible, and enhance things using composition or currying.

Here are my 2 cents:


// Helpers

const pipe = (...fns) => x => fns.reduce((y, f) => f(y), x);
const reverseText = (text = '')=> text.split('').reverse().join('');

const getInitialsDelimitBy = (delimiter = ' ') => (displayName = '') =>
  displayName
    .trim()
    .split(delimiter)
    .reduce((acc, value) => `${acc}${value.charAt(0)}`, '')
    .toUpperCase();

const getInitialsDelimitByComas = pipe(
  getInitialsDelimitBy(','), 
  reverseText
);

const getInitialsDelimitBySpaces = getInitialsDelimitBy(' '); // Not necessary because of the default but clearer 

const comaInitials = getInitialsDelimitByComas('Wayne, Bruce') // BW
const spaceInitials = getInitialsDelimitBySpaces('Bruce Wayne') // BW

For your specific case I would suggest something like this:

const pipe = (...fns) => x => fns.reduce((y, f) => f(y), x);

const nameProcessor = {
  single: (name = '') =>
    name
      .trim()
      .substring(0, 2)
      .toUpperCase(),
  multiple: pipe(
    name => name.trim().split(' '),
    words => `${words[0].charAt(0)}${words[words.length - 1].charAt(0)}`,
    initials => initials.toUpperCase(),
  ),
};

const getInitials = (displayName = '') => 
  displayName.split(' ').length === 1 
    ? nameProcessor.single(displayName) 
    : nameProcessor.multiple(displayName)

getInitials('FullName') // FU
getInitials('FirstName MiddleName LastName') // FL
getInitials('1stName 2ndName 3rdName 4thName 5thName') // 15

I hope that helps =D

Upvotes: -2

Prabs
Prabs

Reputation: 19

This solution uses Array capabilities, Arrow function and ternary operator to achieve the goal in one line. If name is single word, just take first two chars, but if more, then take 1st chars of first and last names. (thanks omn for reminding single word name use case)

string.trim().split(' ').reduce((acc, cur, idx, arr) => acc + (arr.length > 1 ? (idx == 0 || idx == arr.length - 1 ? cur.substring(0, 1) : '') : cur.substring(0, 2)), '').toUpperCase()

Upvotes: 1

Anjan Biswas
Anjan Biswas

Reputation: 7932

This should work for majority of the cases including middle names and first name only (extension on @njmwas answer).

const initialArr = name.split(" ").map((n)=>n[0]);
const init = (initialArr.length > 1)? `${initialArr[0]}${initialArr[initialArr.length - 1]}` : initialArr[0];
const initials = init.toUpperCase();

Upvotes: 0

Mohammed Salahuddin
Mohammed Salahuddin

Reputation: 26

var personName = "FirstName MiddleName LastName";
var userArray = personName.split(" ");
var initials = [];
if(userArray.length == 1){
 initials.push(userArray[0][0].toUpperCase() + userArray[0][1]).toUpperCase();}
else if(userArray.length > 1){
initials.push(userArray[0][0].toUpperCase() + userArray[userArray.length-1][0].toUpperCase());}
console.log(initials);

Upvotes: 0

GeorgeWL
GeorgeWL

Reputation: 322

Using some es6 functionality:

const testNameString = 'Hello World';
const testNameStringWeird = 'Hello  darkness My  - Óld Friend Nikolaus Koster-Walder ';
const getInitials = nameString =>{
       const regexChar = /\D\w+/
       return nameString
        .trim() //remove trailing spaces
        .split(' ') // splits on spaces
        .filter(word => word.length > 0) // strip out double spaces
        .filter(word => regexChar.test(word)) // strip out special characters
        .map(word => word.substring(0, 1).toUpperCase()) // take first letter from each word and put into array
}
console.log('name:',testNameString,'\n initials:',getInitials(testNameString));
console.log('name:',testNameStringWeird,'\n initials:',getInitials(testNameStringWeird));

Upvotes: -1

deejayy
deejayy

Reputation: 770

const getInitials = name => name
  .replace(/[^A-Za-z0-9À-ÿ ]/ig, '')        // taking care of accented characters as well
  .replace(/ +/ig, ' ')                     // replace multiple spaces to one
  .split(/ /)                               // break the name into parts
  .reduce((acc, item) => acc + item[0], '') // assemble an abbreviation from the parts
  .concat(name.substr(1))                   // what if the name consist only one part
  .concat(name)                             // what if the name is only one character
  .substr(0, 2)                             // get the first two characters an initials
  .toUpperCase();                           // uppercase, but you can format it with CSS as well

console.log(getInitials('A'));
console.log(getInitials('Abcd'));
console.log(getInitials('Abcd Efgh'));
console.log(getInitials('Abcd    Efgh    Ijkl'));
console.log(getInitials('Abcd Efgh Ijkl Mnop'));
console.log(getInitials('Ábcd Éfgh Ijkl Mnop'));
console.log(getInitials('Ábcd - Éfgh Ijkl Mnop'));
console.log(getInitials('Ábcd / # . - , Éfgh Ijkl Mnop'));

Upvotes: 3

Vickky Kesty
Vickky Kesty

Reputation: 11

You can do something like this;

    function initials(name){

      //splits words to array
      var nameArray = name.split(" ");

      var initials = '';

      //if it's a single word, return 1st and 2nd character
      if(nameArray.length === 1) {
        return nameArray[0].charAt(0) + "" +nameArray[0].charAt(1);
      }else{
         initials = nameArray[0].charAt(0);
      }
      //else it's more than one, concat the initials in a loop
      //we've gotten the first word, get the initial of the last word


      //first word
      for (i = (nameArray.length - 1); i < nameArray.length; i++){
        initials += nameArray[i].charAt(0);
      }
     //return capitalized initials
     return initials.toUpperCase();
   }

You can then use the function like so;

  var fullname = 'badmos tobi';
  initials(fullname); //returns BT 

  var surname = 'badmos';
  initials(surname); //returns BA

  var more = 'badmos gbenga mike wale';
  initials(more); //returns BW;

I hope this helps.

Upvotes: 1

Aniket
Aniket

Reputation: 582

'Aniket Kumar Agrawal'.split(' ').map(x => x.charAt(0)).join('').substr(0, 2).toUpperCase()

Upvotes: 6

Emerson Bottero
Emerson Bottero

Reputation: 338

  const getInitials = name => {
    let initials = '';
    name.split(' ').map( subName => initials = initials + subName[0]);
    return initials;
  };

Upvotes: 0

njmwas
njmwas

Reputation: 1231

You can use this shorthand js

"FirstName LastName".split(" ").map((n)=>n[0]).join(".");

To get only First name and Last name you can use this shorthand function

(fullname=>fullname.map((n, i)=>(i==0||i==fullname.length-1)&&n[0]).filter(n=>n).join(""))
("FirstName MiddleName OtherName LastName".split(" "));

Upvotes: 58

Andrea
Andrea

Reputation: 3440

Check the getInitials function below:

var getInitials = function (string) {
    var names = string.split(' '),
        initials = names[0].substring(0, 1).toUpperCase();
    
    if (names.length > 1) {
        initials += names[names.length - 1].substring(0, 1).toUpperCase();
    }
    return initials;
};

console.log(getInitials('FirstName LastName'));
console.log(getInitials('FirstName MiddleName LastName'));
console.log(getInitials('1stName 2ndName 3rdName 4thName 5thName'));

The functions split the input string by spaces:

names = string.split(' '),

Then get the first name, and get the first letter:

initials = names[0].substring(0, 1).toUpperCase();

If there are more then one name, it takes the first letter of the last name (the one in position names.length - 1):

if (names.length > 1) {
    initials += names[names.length - 1].substring(0, 1).toUpperCase();
}

Upvotes: 56

Related Questions