jquery div onclick call function

Question

I need to setup a onclick where if you click on the div with the id of "version" it calls the below script.

html

<div id="version">CONTINUE</div>

js

if (/Android\s+([\d\.]+)/.test(navigator.userAgent)){ 
 var anversion=new Number(RegExp.$1) 
 if (anversion=<3)
  window.location = 'main2.html'
 }
else
 window.location = 'main.html'

Useragent sniffing can be a bad practice.. just an FYI

Barry Chapman
– Barry Chapman

2012-02-07 20:55:56 +00:00
Commented Feb 7, 2012 at 20:55 — Barry Chapman
– Barry Chapman, Commented Feb 7, 2012 at 20:55

iblue · Accepted Answer · 2012-02-07 21:12:43Z

3

$("#version").click(function() {
  // Your code here...
});

For more details see the jQuery API on .click()

EDIT: I also corrected your code. It then looks like this:

if(/Android\s+([\d\.]+)/.test(navigator.userAgent)) { 
  var anversion = parseInt(RegExp.$1);
  if(anversion <= 3) {
    window.location.href = 'main2.html';
  } else {
    window.location.href = 'main.html';
  }
}

edited Feb 7, 2012 at 21:12

answered Feb 7, 2012 at 20:54

iblue

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3 Comments

Blainer Over a year ago

I tried that, but Im not even redirected to main.html on click

iblue Over a year ago

That is, because your code is buggy. Please use window.location.href = 'main.html'; and set appropiate semicolons.

iblue Over a year ago

The code only redirects on Android browsers. Are you running the code on your phone?

Chriseyre2000 · Accepted Answer · 2012-02-07 20:57:45Z

0

Have a look at the following: http://api.jquery.com/click/

answered Feb 7, 2012 at 20:57

Chriseyre2000

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Comments

dfsq · Accepted Answer · 2012-02-07 21:03:06Z

0

Try this:

$("#version").click(function() {
    if (/Android\s+([\d\.]+)/.test(navigator.userAgent)) { 
        window.location.href = Number(RegExp.$1) <= 3 ? 'main2.html' : 'main.html';
    }
});

Your code has errors.

anversion=<3 should be anversion <= 3
new Number(RegExp.$1) should be Number(RegExp.$1) otherwise you create an object instead of converting string to number.

answered Feb 7, 2012 at 21:03

dfsq

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