onmouseover

[marcojalongo]

Hello, this is my code:

$(document).ready(function(){
    $("div.ajax" ).on( "mouseover", function() {
        $(this).tipso({
            background: null,
            useTitle: false,
            ajaxContentUrl : 'myfile.php?var='+$(this).data("id")
        });
    });
}); 

it works but why the tooltip is showed only the second time the mouse is over the div element (with class = ajax)?
Sorry, I'm newbe with jquery...

Thank a lot

[object505]

Hello,
You don't need the mouseover event, because tipso is already doing that.
So you just need to do this:

$(document).ready(function(){        
    $("div.ajax").tipso({
        background: null,
        useTitle: false,
        ajaxContentUrl : 'myfile.php?var='+$(this).data("id")
    });        
}); 

[marcojalongo]

Ok, but the problem is that I have a lot of div with class = ajax;
how I can't retrieve $this data of specific div...?

[marcojalongo]

I mean that your code give me var= undefined....

[object505]

Oh, I see.
Maybe something like this would help

var dataID;
$('div.ajax').on({
  mouseenter: function () {
    dataID = $(this).data('id');
  }
});
$("div.ajax").tipso({
    background: null,
    useTitle: false,
    ajaxContentUrl : 'myfile.php?var='+dataID 
});    

I guess in a future release of the plugin I would have to implement this.
Let me know if this works for you :)

[marcojalongo]

it doesn't work... in console log i see "var=undefined" yet

Have you other suggestion?

thanks again...

[object505]

Well this works http://jsfiddle.net/1kb064w7/
Another thing is to give me the site url and let me check what is going on :)

[marcojalongo]

It's work, but how to apply this example to "ajaxContentUrl" (not static "content") I'm using....? I tried to return from function the url of script (e.g.; return 'myscript.php?var='+dataID') but it doesn't work...

[marcojalongo]

sorry... my application is on local server but use exactly same syntax...

[object505]

The same code will be used as here #10 (comment)

Basically that is the same code as in the jsfiddle. If you can make some test environment somewhere I can help you more :)

[marcojalongo]

Ok; I've make following test environment:

1. http://www.giglione.it/appl/mainscript.php

this is your solution (#10 comment); as you can see, ajax request send back "undefined" instead of values "var1", "var2",.... (in console log, too)

2. http://www.giglione.it/appl/mainscript1.php

this is my "dirty" solution; it is ok, but only on second mouse over.

the ajax script for dynamic tip content (myscript.php) actually simply is:

[object505]

Based on your test environment I've tested this code and it is working.

    jQuery(document).ready(function(){          
        var dataID;
        jQuery('div.ajax').on({
            mouseenter: function () {
                dataID = $(this).data('id');            
            }
        });
        jQuery("div.ajax").tipso({
            useTitle: false,
            ajaxContentUrl: 'myscript.php?var='+dataID,
            onBeforeShow: function(){
                $('div.ajax').tipso('update', 'ajaxContentUrl', 'myscript.php?var='+dataID);
            }
        });              
    });

Let me know if it's works for you :)

PS: On your "dirty" version it works from the second hover because on the first you are calling the plugin for the first time and also you are calling the plugin for each element individually, not using the class .ajax but the element itself ( using "this" ).

[marcojalongo]

Great!
thanks a lot!!