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Home/ Questions/Q 6625093
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T21:43:57+00:00 2026-05-25T21:43:57+00:00

I am having problems calling the function callpage Here is the code were I

  • 0

I am having problems calling the function callpage

Here is the code were I am trying to call the function. I got a label that when click I want to call callpage:

$('.gulemenu label, .payload label').toggle(function(){
    $(this).find('input[type="checkbox"]').prop('checked',true);
    $.(this).callpage();
}, function(){
    $(this).find('input[type="checkbox"]').prop('checked',false);
    $.(this).callpage();
});

$('.gulemenu input, .payload label').bind('click',function(e){ e.stopPropagation(); });

My callpage function:

function callpage() {
    $('#formcontent').empty().html('<p style="margin-top:20px;text-align:center;font-family:verdana;font-size:14px;">Vent venligst, henter webhosts.</p><p style="margin-top:20px;margin-bottom:20px;text-align:center;"><img src="../images/ajax.gif" /></p>');
    var form = $('form#search');
    $.ajax({
        type: form.attr('method'),
        url: form.attr('action'),
        data: form.serialize(),
        success:function(msg){
        $('#formcontent').html(msg);
    }})
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T21:43:58+00:00Added an answer on May 25, 2026 at 9:43 pm
     <script type="text/javascript">
        var slider = $("#mySliderSelector").slider({whatever:'you need'});
    
        $('.gulemenu label, .payload label').click(function(event){
    e.stopPropagation();
           var checkbox = $(this).find(":checkbox"); //child checkbox inputs. you are sure there is only one correct?
           var checked = checkbox.prop("checked");
           checkbox.prop("checked", !checked);
           slider.trigger("slidestop"); //jQuery-ui documentation for slideStop event.
        });
        </script>
    
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