jQuery Forum

i need some help with the menu

Basically, i have 2 div  with css(position:absolute)
First div is it a 100x100, and second 30x30
The second div is hide and is located in the center of th first div.

When the mouse is mouseover de the first div then show the second, and when is leave form first then hide the second div

$('#div1').mouseover(function(){
   $("#div2").show();
});

$('#div1').mouseout(function(){
   $("#div2").hide();
});


it ok, but when i mouve the mouse on the second div then the mouseover property is hide the div and show.
is like a infinite loop.

i want to show the second div until the mouseout the first div in totality.

Thanks, amic.

Hi all,

i want to validate a form with mootools and i use the submit function jquery for ajax response.
if i submit ... the mootools show me the error if exists, is good, but if i have this error jquery submit  passing from them (error)

i want to jQuery submit when the mootools validation is correct.

this is the code

  <script language="javascript" src="jquery-1.3.min.js"></script>
  <script language="javascript">
  jQuery.noConflict();
  jQuery(document).ready(function() {
   // Mostramos el loader
    jQuery().ajaxStart(function() {
        jQuery('#loading').show();
        jQuery('#result').hide();
    }).ajaxStop(function() {
        jQuery('#loading').hide();
        jQuery('#result').fadeIn('slow');
    });
   // Enviamos el formulario
    jQuery('#enviar_anuncio').submit(function() {
   // Definimos el metodo ajax, los datos
        jQuery.ajax({
            type: 'POST',
            url: jQuery(this).attr('action'),
            data: jQuery(this).serialize(),
            success: function(data) {
                // Imprimimos la respuesta en el div result
                jQuery('#result').html(data);
            }
        })
        return false;
    });
  }) 
  </script>


     <script type="text/javascript" src="js/formcheck/mootools.js"></script>
     <script type="text/javascript" src="js/formcheck/formcheck.js"></script>
     <script type="text/javascript">
      window.addEvent('domready', function(){
       formcheck = new FormCheck('enviar_anuncio', {
         display : {
                 fadeDuration : 500,
                 errorsLocation : 1,
                 indicateErrors : 1,
           keepFocusOnError : 1,
           flashTips  : 1,
           removeClassErrorOnTipClosure  : 1,
                 showErrors : 0
             }     
        })
     });
     </script>

Thanks, amic.

Hi all,


If i submit the form .... return the "n" results with pagination ( 10 for page).

And i have  5(for example) page with 10 result for page .

This is work fine, but in me database i have more rows (500 for example), and if i press the second(2) button of the pagination (2 page ) then return  me all rows of the database because she lost the selection of the form(input, radio buttons, select ...).

How to send the values of the form when click on the 2, 3, ... page of the pagination link.

Sorry for me bad english.
Thanks, amic.

Hi all..

I work in this page

http://tibuleac.ro/sondaj/index/1/Mihai_Tabuleac

you can see in the left are a slide with jQuery

this is ok is work fine, but if you go to the bottom of the page and wait to change the slider you can see how the page is moving on the middle(or top).

if you go a page  http://tibuleac.ro/sustinatori/index
and the bottom, wait to change... upss he don´t change .. because the right part of the page not empty.

I dont want the page to move.
Help me pls..

Hi all,

i have some problem with me code, please help.

i have a select  named categoria:

1.     $('categoria').addEvent('change', function(){
   
          var selection = document.getElementById('categoria').value;     
2.      
   
3.        formcheck.dispose($('ano')); 
4.        formcheck.dispose($('titulo'));
5.        formcheck.dispose($('marca'));
6.        formcheck.dispose($('version'));
7. 
   
8.        if(selection==1){
             formcheck.register($('ano')); 
          }
          if(selection==2 || selection==3 || selection==6 || selection==7)  {
             formcheck.register($('titulo'));
          }
          if(selection==4){
             formcheck.register($('marca'));
             formcheck.register($('version'));
           } 
9.  });
   

This is work perfectly, but i have the selection var already defined(sometimes) and  if  the selection var exists
and it`s value is 4 then activate the selection for number 4  (form check.register($('marca')); formcheck.register($marca('$version'));)

Thanks, amic.

Hi all,

how to get a value of a div when this change value?

example

<div id="count"> 0 </div>

Put value:

$('#count').val()=3

jQuery('#count').change(function() {
 alert(jQuery(this).val());
}

Thanks, amic.

Hi all,

i have this code:

          //delete image
          var limit= 5;
          var count = <?php echo count($anuncio_fotos); ?>
          <?php
          foreach($anuncio_fotos as $item):
          ?>
          var name = '<?php echo $item['name']; ?>';
          var li = jQuery('<li><img src="' + base_url + 'uploads/anuncios/thumbs/'+ name +
                        '" width="100" alt="' + base_url + 'uploads/anuncios/thumbs/'+ name +
                        '" title="' + base_url + 'uploads/anuncios/thumbs/'+ name +
                        '"/><br /><div class="delete_image_upload">Eliminar</div></li>');
                    li.hide().insertBefore('#displayFiles').slideDown('slow').find('.delete_image_upload').click(function () {
                        li.fadeOut('slow')
                        count-=1;
                        if(count<limit) { jQuery('.well').show(); }
            //urmatoarele linie este pentru a ascunde div de la //recuerda que con fotos tu anuncio sera mas visible
            if(count==0) { jQuery('div.text_upload2').show().slideDown('fast');  }
                        jQuery.ajax({
                            url: base_url + 'uploadify/delete_file2/'+name+'/<?php echo $item['id']; ?>',
                            success: function (response) {
                                li.remove()
                            }
                        });
            return false
                  });
          <?php
          endforeach;
          ?>

and i want to delete a image when click Eliminar en div delete_image_upload

but, if i have 5 images and i want to delete the second image .. the click delete the last image from div displayFiles and the corectly from folder uploads...

Help me.. Thanks, amic.