JavaScript/ajax

Ajax - Ajax.Request

아이티.파머 2010. 8. 17. 19:31
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new Ajax.Request(actionURL,{
   method: 'post',
   parameters : {companyName : $F('coporatedCompany')},
   onLoading  : function(){},
   onSuccess  : function(response){
        var resultObj = eval('(' + response.responseText + ')');
        var resultValue = resultObj.result;
        if ( resultValue == "SUCCESS" ) {
            var tbody = $('tb1').select('tbody')[0];
              while(tbody.hasChildNodes()){
                     tbody.removeChild(tbody.lastChild);
             }
            var companyList = resultObj.companyList;
            for(var i = 0 ; i < companyList.length ; i++){
      
            };
          }
       },
   onFailure  : function(){alert('뿅');}
  });

actionURL 이곳에는,  URL 경로를 적어 주면 되겠지,
 

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