Question

I am trying to create an array with different hrefs to then attach to 5 separate elements.

This is my code:

var link = new Array('link1', 'link2', 'link3', 'link4', 'link5');

                    $(document.createElement("li"))
                    .attr('class',options.numericId + (i+1))
                    .html('<a rel='+ i +' href=\"page.php# + 'link'\">'+ '</a>')
                    .appendTo($("."+ options.numericId))

As you can see I am trying to append these items from the array to the end of my page so each link will take the user to a different section of the page. But i have not been able to do this. Is there a way to to create elements with different links?

I am new to javascript so I am sorry if this doesn’t make a whole lot of sense. If anyone is confused by what i am asking here I can try to clarify if I get some feedback.

The code I would Like output is:

    <ul class="controls">
    <li class="controls1"><a href="page.php#link1"></a></li>
    <li class="controls2"><a href="page.php#link2"></a></li>
    <li class="controls3"><a href="page.php#link3"></a></li>
    <li class="controls4"><a href="page.php#link4"></a></li>
    <li class="controls5"><a href="page.php#link5"></a></li>
    </ul>

Which is similar to what I am getting, however when I apply the fix that andres descalzo has supplied, my list elements are each repeating themselves 5 times.

Any solutions would be greatly appreciated.

Thanks,

jason

Answer 1

something like this?:

*Edit II * for comment

var link = ['strategy', 'branding', 'marketing', 'media', 'management'],
          refNumericId = $("."+ numericId);

  $(link).each(function(i, el){

      $("<li></li>")
            .attr("id", numericId + "_" + (i+1))
            .attr("class", numericId + (i+1))
            .html("<a href=\"capabilities.php#"+el+"\"></a>")
            .appendTo(refNumericId);

  });

I saw your code in the file ‘easySlider1.7.js’ and you’re including in ‘for’ of the line 123 the code ‘var link = [‘ strategy,”which should go after this ‘for’