Syntax to run a php Function inside javascript

Question

I have tried different approaches to get this to work. I know I have to wrap the function but can not get it to work. The function is running in a .js file.

http://jsfiddle.net/hookedonweb/yq8Hz/

function showLabel(labelName) {
   var labelImgs = jQuery(".content .packaging .labels img");
   var focusImg  = jQuery("#label_" + labelName);

   if (labelImgs.length > 0 && focusImg.length > 0) {
       labelImgs.each(function() { jQuery(this).removeClass("active"); });

       focusImg.addClass("active");
   }
}

Given your function works.. the only things I can assume are that you are not including jquery in your page, and that your function is not in the document head. — Daedalus
– Daedalus, Commented Nov 3, 2012 at 5:19

mrtsherman · Accepted Answer · 2012-11-03 05:22:07Z

1

I think you may be confusing PHP and jQuery. Anyways, your code can be simplified a lot. This seems to do what you want. If it doesn't then you need to clarify your question.

http://jsfiddle.net/yq8Hz/3/

Don't use inline javascript. You are using jQuery, so use it to separate your code from your html. This is a good practice. You can use the data attributes to store the img you want to reference.

<div><a href="" data-label="macandcheese">See Label</a></div>
<div><a href="" data-label="lasagna">See Label</a></div>
<div><a href="" data-label="beefstroganoff">See Label</a></div>

Then use jQuery's hover to add/remove the active class. We make use of jQuery's data method to get the label we want.

$('a').hover( function() {
    $('img').removeClass('active');
    $('img#label_' +  $(this).data('label')).addClass('active');
});

answered Nov 3, 2012 at 5:22

mrtsherman

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