Using jQuery to copy value from image source's file name to process and paste into html?

Question

Hi i'm trying to use jQuery/javascript to make my life easier. I'm trying to use it to auto fill in the various html elements with the name of the place and country which comes from the file name of the image. Below is a sample of some of the html i have to work with:

<div class="boxgrid">
<a href="gallery/angkor - cambodia.jpg" title="[PLACE] - [COUNTRY]">
  <img src="gallery/thumbs/angkor - cambodia.jpg" alt="[PLACE] - [COUNTRY]" width="100" height="100"/>
</a>
  <div class="boxcaption">
    <span class="place">[PLACE]</span>
    <span class="country">[COUNTRY]</span>
  </div>
</div>

Imagine this block of html is repeated multiple times for every image in the gallery so i assume .each() will need to be used.

The idea is that [NAME OF PLACE] and [NAME OF COUNTRY] should be filled in with the image's file name using the - (dash) as the delimiter between the two values. How can I accomplish this with jQuery/javascript? (i realise i should use a server side language or go about this another way but this is what i'm restricted to)

As a bonus the first letter of each of the values should be capitalised, for example Angkor - Cambodia .

Thank you, any help would be much appreciated.

Answer 1

Here's the full code with trimming and capitalization of the first letter for Place and Country.

String.prototype.capitalize = function(){
    return this.replace(/\S+/g, function(a){
        return a.charAt(0).toUpperCase() + a.slice(1).toLowerCase();
    });
};

$(function(){
 $('.boxgrid').each(function(){
    var placeCountry = $(this).find('img').attr('src')
                            .replace('gallery/thumbs/','')
                            .replace('.jpg','').capitalize();

    var place = $.trim(placeCountry.split('-')[0]); 
    var country = $.trim(placeCountry.split('-')[1]);
    $(this).find('a').attr('title',placeCountry);
    $(this).find('img').attr('alt',placeCountry);
    $(this).find('.place').html(place);
    $(this).find('.country').html(country);
 });
});