Get all string in one variable after space in javascript

Question

what i am doing

for example i have a name

 var nameofuser = 'John Constantine vayo';
    var values = nameofuser.split(" ");
    var f_name = values[0];
    var l_name = nameofuser.substr(nameofuser.indexOf(' ') + 1);
    console.log(f_name);
    console.log(l_name);

it is working fine when user entering his last name and other title or so on, it is all going in l_name but when a user enterned only first name like only 'John' then l_name is also populated with 'john' :(,

i want IF a user type last name and so on then only l_name should be populated.

if user type only first name then only f_name with first name should be populated

if a user type last name and so on then only l_name should be populated. if user type only first name then only f_name with first name should be populated — Aman
– Aman, Commented Nov 24, 2016 at 11:17
It sounds like you want a script that recognizes whether a single word is a first name or a last name. How is it supposed to do that? — user5734311
– user5734311, Commented Nov 24, 2016 at 11:17

Gaurav joshi · Accepted Answer · 2016-11-24 11:17:43Z

6

var nameofuser = 'John Constantine vayo';
var values = nameofuser.split(" ");
var f_name = values[0];
var l_name = values[1] ? nameofuser.substr(nameofuser.indexOf(' ') + 1) : '';
console.log(f_name);
console.log(l_name);

answered Nov 24, 2016 at 11:17

Gaurav joshi

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2 Comments

Gaurav joshi Over a year ago

If it works then please upvote my answer and check this as right answer @adasdasd

Rajesh Dixit Over a year ago

@Gauravjoshi You can look into values.join(' ') for l_name

Denys Séguret · Accepted Answer · 2016-11-24 11:23:37Z

1

Regular expressions are the usual tool for this kind of decomposition:

var nameofuser = str.match(/^\s*(\S+)\s*(.*?)\s*$/).slice(1);

'John Constantine vayo' gives ["John", "Constantine vayo"]

and 'John' gives ["John"].

But be careful of this kind of guesses. In a multicultural world it's safer to just take what the user provides.

edited Nov 24, 2016 at 11:23

answered Nov 24, 2016 at 11:18

Denys Séguret

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2 Comments

alebianco Over a year ago

I approve of this, but it might break if the string begins with a space. You just need to trim the string before matching it.

Denys Séguret Over a year ago

@alebianco I tuned the regex to do the trimming

Rajaprabhu Aravindasamy · Accepted Answer · 2016-11-24 11:38:02Z

1

When searching for a space over a string that doesn't have a space -1 would be returned. So -1 + 1 will be evaluated to 0 and the substring is returning the entire string.

var nameofuser = 'John Constantine vayo';
var values = nameofuser.split(" ");
var f_name = values.shift();
var l_name = values.length ? values.join(" ") : undefined;
console.log(f_name);
console.log(l_name);

edited Nov 24, 2016 at 11:38

answered Nov 24, 2016 at 11:19

Rajaprabhu Aravindasamy

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3 Comments

Rajesh Dixit Over a year ago

You can skip ternary operation. Split will always return you an array of at least 1 value. Just use values.shift() for first name, and values.join(' ') for lastname

Rajesh Dixit Over a year ago

This will give you John and Constantine only. l_name should be Constantine vayo. You can check my answer, as I have already done that.

Rajaprabhu Aravindasamy Over a year ago

@Rajesh Yeah. I hope the current edit would address the issue that you were talking about.

malutki5200 · Accepted Answer · 2016-11-24 11:19:05Z

0

It Is because You have to make sure there is more than one word for that line:

var l_name = nameofuser.substr(nameofuser.indexOf(' ') + 1);

Your code should look something like this:

var nameofuser = 'John Constantine vayo';
var values = nameofuser.split(" ");
var f_name = values[0];
var l_name = (values.length > 1) ? nameofuser.substr(nameofuser.indexOf(' ') + 1) : '';
console.log(f_name);
console.log(l_name);

answered Nov 24, 2016 at 11:19

malutki5200

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Comments

Rajesh Dixit · Accepted Answer · 2016-11-24 11:26:32Z

0

Just check if there is more than one word in the values array? Hope this helps :)

var values = nameofuser.split(" ");
var f_name = values[0];
if (values.lenght >= 2) {
  var l_name = nameofuser.substr(nameofuser.indexOf(' ') + 1);
} else { //let lastname empty or do something else}

  console.log(f_name);
  console.log(l_name);
}

edited Nov 24, 2016 at 11:26

Rajesh Dixit

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answered Nov 24, 2016 at 11:23

Ruckert Solutions

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

When you do .indexOf(), if value is not matches, you will get index as -1 so indexOf() + 1 would interpret as -1 + 1 = 0. So you end up doing namesofuser.substring(0)

Try something like this:

string.split + array.shift and array.join

function getNames(str) {
  var values = str.split(" ");
  var f_name = values.shift();
  var l_name = values.join(' ');
  console.log(f_name);
  console.log(l_name);
}

getNames('John Constantine vayo')
getNames("Foo")

string.substring

function getNames(str) {
  var index = str.indexOf(' ');
  index = index > -1 ? index : str.length;
  var f_name = str.substring(0, index);
  var l_name = str.substring(index + 1);
  console.log(f_name);
  console.log(l_name);
}

getNames('John Constantine vayo')
getNames("Foo")

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Nov 24, 2016 at 11:21

Rajesh Dixit

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Comments

Abinesh Joyel · Accepted Answer · 2016-11-24 11:32:42Z

0

var nameofuser = 'john';
var f_name = '';
var l_name ='';
var values = nameofuser.split(" ");
var f_name = values[0];
if(values.length > 1){
     var l_name = nameofuser.substr(nameofuser.indexOf(' ') +1);
  }       
console.log(f_name);
console.log(l_name);

edited Nov 24, 2016 at 11:32

answered Nov 24, 2016 at 11:26

Abinesh Joyel

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Comments

alebianco · Accepted Answer · 2016-11-24 11:41:12Z

0

this works as well and fixes the problems you'll have if nameofuser begins with a space

var nameofuser = 'John Constantine vayo';
var values = nameofuser.trim().split(" ");
var f_name = values[0];
var l_name = values.slice(1).join(" ")
console.log(f_name);
console.log(l_name);

answered Nov 24, 2016 at 11:41

alebianco

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