小开

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/

小开

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();
}

小开

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);

小开

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('');

小开

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(" "));

小开

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("");

小开

just updated Andrea's version:

var getInitials = function (string) {
var initials = "";
var names = string.split(' ');
for (n = 0; n < names.length; n++) {
initials += names[n].substring(0, 1).toUpperCase();
}
return initials;
};

if string includes LastName, just change names.length to names.length-1 to ignore lastname

小开

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

小开

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

小开

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.

小开

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'));

小开

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));

小开

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"

小开

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);

小开

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();

小开

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()

小开

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

小开

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.

小开

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

小开

Easy way using ES6 Destructering:

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

小开

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

小开

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

小开

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();
}

小开

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();
}

小开

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

小开

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.

小开

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("") || ""
);
}

小开

+ 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')

小开

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'));

小开

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();
}

小开

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('')

Getting Name initials using JS

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

Common Avatar Use-case

Note