Javascript - catch name in variable string

Question

I'm trying to catch a person's name. The name would be entered in a text box such as:

my name is Robert or yes my name is Robert etc.

I don't know where the actual name will fall however because of intro words etc.

I was thinking something like this.

I search for "my name is"
I capture it in an array
I split the array
I now know the actual name follows as such:

namesParts[0] - would be "my"

namesParts[1] - would be "name"

namesParts[2] - would be "is"

namesParts[3] - would be the name i'm looking for.

Something perhaps like the below but this doesn't work.

if (input.search("my name is")!= -1) {
    var names = input.match(/my name is/);
    var namesParts = names.split(' '); 
    var one = namesParts[3];
    document.result.result.value = "Ok your name is "+one+".";
    return true;
}

"Robert is my name." "I'm known as Robert." "Call me Bruce." "Robert Peters." — nnnnnn
– nnnnnn, Commented May 12, 2013 at 21:49

zerkms · Accepted Answer · 2013-05-12 21:42:21Z

1

If all other words will start with lower case letter you could use

'my name is Robert'.match(/[A-Z]+\w*/);

otherwise

'My name is Robert'.match(/my name is (\S+)/i);

answered May 12, 2013 at 21:42

zerkms

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3 Comments

mister movie Over a year ago

ok once I got the match then something like this: var names = names[1];

zerkms Over a year ago

@mister movie: I'm still not sure what you're talking about

mister movie Over a year ago

I was trying to fit your example into my original "if statement" presented above. I think I figured it out. Thanks!

jmingov · Accepted Answer · 2013-05-12 22:15:58Z

0

Check the JavaScript String.split Method.

Examples:

var str="my name is Robert";
var n=str.split('my name is ');
alert('1. Length: '+n.length +' Array: '+ n);

var str="my name is Robert";
var n=str.split(' ');
alert('2. Length: '+n.length +' Array: '+ n);

var str="my name is Robert";
var n=str.split('');
alert('3. Length: '+n.length +' Array: '+ n);

Live example: http://jsfiddle.net/a4D8q/

answered May 12, 2013 at 22:15

jmingov

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