Reputation: 3
I am trying to utilize Pinterest's pin it button on my website. The directions on the Pinterest 'create a button' section are straight forward except for the fact that I'm only given the option of entering the url of the image I want pinned. This can be very strenuous and redundant, especially for someone working on a website that has more than a hundred pages (pages containing product images). So I figured I'd use some Javascript (jQuery preferably) to do the trick.
Here's my code.
HTML:
<div id="pinterest"></div>
<!--Sample Image-->
<div id="pic">
<img id="sampleimage4div" src="images/wonderfulpicture.jpg" />
</div>
<!--SCRIPTS-->
<script type="text/javascript" src="http://assets.pinterest.com....
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/... etc
jQuery Script
<script type="text/javascript">
$(document).ready(function(){
var currentURL = $(location).attr('href');
var description = encodeURIComponent($(this).attr('title'));
var imageURL = encodeURIComponent($('#sampleimage4div').attr('src')+currentURL);
$('#pinterest').append('<a href="http://pinterest.com/pin/create/button/?url='+currentURL+'&media='+imageURL+'&description='+description+'"class="pin-it-button"count-layout="horizontal"><img border="0" src="http://assets.pinterest.com/images/PinExt.png"title="Pin It"/></a>');
});
</script>
I was able to alter/customize the page 'URL' and 'description' but for some reason I'm failing to get the image.
Any ideas?
Upvotes: 0
Views: 838
Reputation: 16959
var currentURL = $(location).attr('href');
var imageURL = encodeURIComponent($('#sampleimage4div').attr('src')+currentURL);
Those lines of code are the problem. use
var currentURL = $(location).attr('host')
var imageURL = encodeURIComponent("//" + currentURL + $('#sampleimage4div').attr('src'));
Upvotes: 1
Reputation: 210
once use this:
$('#pinterest').append('<a href="anything"><img src='+imgvariable+'></a>');
also you can use here:
$('#sampleimage4div').attr('src')+currentURL //at the place of this
use
$('#sampleimage4div').attr('src','src_which_you_want')
Upvotes: 0