JQuery UI對話框不會彈出

Question

我嘗試構建一個公告彈出功能。目前，我有一個回叫函數來檢查是否有人修改announcement.html。

所以這裏是代碼

<div id="popupDialog" class="cs_AnnoucementPop"> 
    <?=$this->data['Popup']?> 
    test 
</div> 

<script type="text/javascript"> 
setInterval("_checkUS2PopUpUpdate()", 12000); //12 seconds (debugging - change back to 50) 
</script> 

所以，每12秒，就有一個Ajax調用

function _checkUS2PopUpUpdate() 
{ 
    var callback=new Object(); 
    callback.success=this.onExternalSuccess; 
    callback.failure=this.onExternalFailure; 
    YAHOO.util.Connect.asyncRequest('GET','/ci/ajaxCustom/ajaxUS2CheckPopupUpdate',callback); 
}; 
function onExternalSuccess (o){ 
    if(o.responseText!==undefined) 
    { 
    var str=o.responseText; 
    //document.getElementById('updateContent').innerHTML=str; 
    if(str !== 'no update2') // Then pop up. 
    { 
     // This code is valid for jQuery-ui but it is not working. I think jQuery is being imported somewhere else and overwriting 
     // jQuery-ui - I have set it back to the original popup for now - Matt Drewery 
      $.noConflict(); 
     $.get('/euf/assets/announcements/pop_up_announcement_us2.html', function(data) { 
        $('#popupDialog').html(data); 
        $('#popupDialog').dialog({modal: true}); 
      }); 

      /*L=screen.width-200; 
     T=screen.height; 
     popup=window.open(str,"",'alwaysRaised=yes,status=no,toolbar=no,location=no,menubar=no,directories=no,resizable=no,scrollbars=no,height=200,width=330,left="+L+",top="+T'); 
     for (i=0;i<200;i++) 
     { 
     T=T-1; 
     popup.moveTo(L,T); 
     }*/ 
     } 
    } 

} 

當有人發消息，它調用加載HTML文件，並嘗試通過使用JQuery UI對話框彈出

但是，我得到了一個突破錯誤。我不知道如何走得更遠。謝謝。

$( [Break On This Error] $('#popupDialog').dialog({modal: true});

Answer 1

使用jQuery，你不能使用noConflict方法，後來使用$符號。檢查這個文檔上http://api.jquery.com/jQuery.noConflict/

試試這個

$.noConflict(); 
jQuery.get('/euf/assets/announcements/pop_up_announcement_us2.html', function(data) { 
jQuery('#popupDialog').html(data); 
jQuery('#popupDialog').dialog({modal: true}); 
});