jQuery rewrite href

Question

I have an onclick function that I want to add an anchor to the href value. I do not want to change the URL's, because I need the site to still function for people without javascript / for SEO purposes. so here is what I have tried using (amoung other things):

jQuery('a[rel=ajax]').click(function () {
    jQuery(this).attr('href', '/#' + jQuery('a'));
});

An orginal link looks like this:

http://www.mysite.com/PopularTags/

URL re-write should look like this, so that AJAX will work:

http://www.mysite.com/#PopularTags

I was able to get some URL's to work by setting a links name value to the same as the href, but it did not work for links with sub-sections:

http://www.mysite.com/Artist/BandName/

So not really sure. Thanks for the help.

Answer 1

I believe something like

jQuery(function(){
  jQuery('a').each(function(){
    jQuery(this).attr('href','/#'+jQuery(this).attr('href'));
  });
});

will do the job. Read as: At page load, loop through all your links, and replace their href attributes with #{old href}.

Answer 2

It depends how you code your links in the HTML. If you’re linking to http://www.mysite.com/PopularTags/ using this:

<a href="/PopularTags/">Popular Tags</a>

You can simply use this:

$(function() {
 $('a[rel=ajax]').each(function() {
  var actualHref = $(this).attr('href');
  this.href = '/#' + actualHref.substring(1, actualHref.length - 1); // cuts off the first and last characters (‘/’)
 });
});

This will result in the following HTML:

<a href="/#PopularTags">Popular Tags</a>

Note that we can set this.href instead of $(this).attr('href'), but we still need the latter to get the actual href attribute. this.href always holds a full URI, including protocol, domain name, etc — which we don’t want in this example.

Answer 3

I'd suggest to use regular expressions, like

$('a[rel=ajax]').click(function(){
  $(this).attr('href', function(){
    this.href = this.href.replace(/\/$/, "");
    return(this.href.replace(/(.*\/)/, "$1#"));
  });
});

Kind Regards

--Andy