jQuery获取json后使用zy_tmpl生成下拉菜单_php技巧

<!--下拉列表开始-->
<div class=" ui-has-label">
  <label class="ui-select ">选择楼群：</label>
  <div class="ui-select">
    <div class="ui-btn ui-btn-icon-right ui-btn-corner-all ui-btn-b">
      <span class="ui-btn-inner ui-btn-corner-all" >
        <span class="ui-btn-text">楼名</span>
        <div class="ui-li-link-alt ui-btn ui-btn-corner-right ui-shadow">
        <span class="ui-icon ui-icon-arrow-d ui-icon-shadow"></span>
        </div>
      </span>
      <select name="lc" id="lc" selectedIndex="0" id="select-choice-0" onchange="zy_slectmenu(this.id)">
        数据加载中，请稍候
      </select>
    </div>
  </div>
</div>
<!--下拉列表结束-->

function getlq(){
  $.getJSON('http://localhost/json.php?jsoncallback=?', function(data) {
    var tmpl='<option value="${BuildingId}">${BuildingName}</option>';
    $("#lc").html("");
    $("#lc").append(zy_tmpl(tmpl,data,zy_tmpl_count(data)));
  })
}