~~~
[javascript]
<script>
$.fn.googlemap = function(){
        var z = $(this);
 
         // author: Christian Salazar, <christiansalazarh@gmail.com>          var src='';
 
$(this).each(function(){
 
var z = $(this);
 
var address = jQuery.trim(z.attr('streetnumber'))                  +'+'+jQuery.trim(z.attr('streetname'))                  +'+'+jQuery.trim(z.attr('cityname'))                  +'+'+jQuery.trim(z.attr('statecode'))                  +'+'+jQuery.trim(z.attr('zipcode'))         ;
 
        var  ;
 
src="https://maps.google.com/maps?"                  +"client=safari"                 +"&q="+address
 
                 +"&q="+address
 
+"&oe=UTF-8&ie=UTF8&hq="                  +"&hnear="+address                 +"&gl=us"
 
                 +"&gl=us"
 
+"&z="+z.attr('zoom')                  +"&output=embed";          z.html("<iframe src='"+src+"' width="+z.attr('width')+" height="                  +z.attr('height')+"></iframe>");          });
 
return src;
}
</script>[javascript]
<script>$('#map').googlemap();</script>
// please dontyou can use it forin multiple selectors, // as an example: $("ul.maps li").googlemap();
 
// will process all LI items having the required attributes
 
// displaying a map each.
 
~~~

It will produce: