jquery ajax select解决方法

jquery ajax select

$.ajax({
					  type: 'POST',
					  url: "${base}/smallLine/findSmallByBigLineId.do",
					  data: "bigLineId="+bigLine,
					  success: function(data){
						  $("#smallLineId").empty();
						  $("#smallLineId")("");
						  var options = "";
						  for ( var i = 0; i < data.smallLinesList.length; i++) {
							  var smallLine = data.smallLinesList[i];
							  options+="<option value='"+smallLine.id+"'>"+smallLine.name+"</option>";
						  }
						  $("#smallLineId").append(options);
						  $("#smallLineId")(options);
						  $("#smallLine").attr("style","");
					  },
					  dataType: "json"
					});

------最佳解决方案--------------------------------------------------------
success: function(data){
                          $("#smallLineId").empty();
                        //  $("#smallLineId")("");
                          var options = "";
                          for ( var i = 0; i < data.smallLinesList.length; i++) {
                              var smallLine = data.smallLinesList[i];
                              options="<option value='"+smallLine.id+"'>"+smallLine.name+"</option>";
                           $("#smallLineId").append(options);
                          }
                         
                         // $("#smallLineId")(options);
                          $("#smallLine").attr("style","");
                      },
------其他解决方案--------------------------------------------------------

var options = "";
for ( var i = 0; i < data.smallLinesList.length; i++) {
 var smallLine = data.smallLinesList[i];
  options+="<option value='"+smallLine.id+"'>"+smallLine.name+"</option>";