Extracting a URL parameter with JavaScript [duplicate]

Question

I have a URL:

http://www.youtube.com/watch?v=JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382

Edit: I should also not the url is stored in a variable and I want it to work something like this:

$(".videothumb a").live('click', function() {
        var URL = < do something to cut the string > 
        console.log(URL);
        return false;
    });

And I want to cut the URL starting from "=" and ending at "&" so I'll end up with a string like this: "JssO4oLBm2s".

I only know of the slice() function but I believe that only takes a number as beginning and end points.

Sorry, my mistake I just updated.

UzumakiDev
– UzumakiDev

2013-11-09 01:31:39 +00:00
Commented Nov 9, 2013 at 1:31 — UzumakiDev
– UzumakiDev, Commented Nov 9, 2013 at 1:31
did you try the answers posted below?

Venkata Krishna
– Venkata Krishna

2013-11-09 01:32:45 +00:00
Commented Nov 9, 2013 at 1:32 — Venkata Krishna
– Venkata Krishna, Commented Nov 9, 2013 at 1:32
@KyleJoseph see me answer with fiddle demo

Tushar Gupta - curioustushar
– Tushar Gupta - curioustushar

2013-11-09 01:51:11 +00:00
Commented Nov 9, 2013 at 1:51 — Tushar Gupta - curioustushar
– Tushar Gupta - curioustushar, Commented Nov 9, 2013 at 1:51

Arun P Johny · Accepted Answer · 2013-11-09 01:33:56Z

1

Using .split() will give a position based solution which will fail the order of parameters changes. Instead I think what you are looking for is the value of parameter called v for that you can use a simple regex like

'http://www.youtube.com/watch?v=JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382'.match('[?&]v=(.*?)(&|$)')[1]

edited Nov 9, 2013 at 1:33

answered Nov 9, 2013 at 1:22

Arun P Johny

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5 Comments

Arun P Johny Over a year ago

@Krishna updated to fetch value of v

UzumakiDev Over a year ago

My url is stored in a variable so this doesn't work.

Arun P Johny Over a year ago

@KyleJoseph then use the variable like myvar.match('[?&]v=(.*?)(&|$)')[1]

UzumakiDev Over a year ago

I did and got this error "Uncaught TypeError: Cannot read property '1' of null "

Arun P Johny Over a year ago

@KyleJoseph it looks like the url don't have the parameter v=?

Oriol · Accepted Answer · 2013-11-09 01:36:36Z

1

Try

'http://www.youtube.com/watch?v=JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382'
    .split('=')[1] // 'JssO4oLBm2s&list'
    .split('&')[0] // 'JssO4oLBm2s'

Or, if you want to be sure to get the v parameter,

var v, args = 'http://www.youtube.com/watch?v=JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382'.split("?")[1].split('&');
for(var i = args.length-1; i>=0; --i) {
    var data = args[i].split('=');
    if(data[0]==='v') {  v = data[1]; break;  }
}

edited Nov 9, 2013 at 1:36

answered Nov 9, 2013 at 1:23

Oriol

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Comments

Venkata Krishna · Accepted Answer · 2013-11-09 01:22:09Z

0

Use .split(). Separated to 2 lines for clarity

var first = "http://www.youtube.com/watch?v=JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382".split('=')[1]
var result = first.split('&')[0]; //result - JssO4oLBm2s

answered Nov 9, 2013 at 1:22

Venkata Krishna

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Comments

Oriol · Accepted Answer · 2013-11-09 01:38:32Z

0

v = 'JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382';

var vamploc = v.indexOf("&");

vstring = v.substr(0, vamploc);

You can play with the code a bit to refine it, but the general concepts work.

edited Nov 9, 2013 at 1:38

Oriol

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answered Nov 9, 2013 at 1:32

TimSPQR

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Comments

Tushar Gupta - curioustushar · Accepted Answer · 2013-11-09 01:48:38Z

0

use Regexp (?:v=)(.+?)(?:&|$)

Fiddle DEMO

"http://www.youtube.com/watch?v=JssO4oLBm2s&list=PLGHJ4fVazTpYRZTEhqgurtSH6XlDMIEJM&shuffle=382".match('(?:v=)(.+?)(?:&|$)')[1]

Reference

http://gskinner.com/RegExr/

answered Nov 9, 2013 at 1:48

Tushar Gupta - curioustushar

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