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);

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(" ")); 

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

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(); } 

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

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(""); 

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

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"

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

+ 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.includes(' ') return name.substring(0, hasTokens ? 1 : 2) + (hasTokens ? name.charAt(name.lastIndexOf(' ') + 1) : '') } console.log(getInitials("Jan Czochralski"), 'JC') console.log(getInitials("A B"), 'AB') console.log(getInitials('SXyz '), 'SX')

You can do a function for that:

var name = 'Name'; function getInitials( name,delimeter ) { if( name ) { var array = name.split( delimeter ); switch ( array.length ) { case 1: return array[0].charAt(0).toUpperCase(); break; default: return array[0].charAt(0).toUpperCase() + array[ array.length -1 ].charAt(0).toUpperCase(); } } return false; } 

Fiddle: http://jsfiddle.net/5v3n2f95/1/

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')); 

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. 

Easier with map function:

var name = "First Last"; var initials = Array.prototype.map.call(name.split(" "), function(x){ return x.substring(0,1).toUpperCase();}).join(''); 

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(); } 

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

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.

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() 

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.

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("") || "" ); } 

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

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

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); 

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(); 

Easy way using ES6 Destructering:

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

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(); } 

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 

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'));

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(); } 

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('')