Home > Web Front-end > JS Tutorial > body text

用jquery生成二级菜单的实例代码_jquery

WBOY
Release: 2016-05-16 17:31:36
Original
970 people have browsed it

javascript

页面中的函数

复制代码 代码如下:

function querySub(id,sid/*选中的项*/){
$.post('${path}/page/article/querySubChannelArticle.action',
                {'channelId':id},function(data){
                    $('select[name="article.subChannel.subId"] option[value!=""]').remove();
                    var jsonObj = data;   
                    for ( var i = 0; i                         var $option = $('

页面代码:
复制代码 代码如下:


                    一级栏目
                         
                           
                         
                
                
                    二级栏目
                         
                         
                         
                

服务器端

复制代码 代码如下:

 public String querySubChannelArticle(){ 
        HttpServletResponse response=getResponse();
        HttpServletRequest request=getRequest();
        response.setContentType("application/json; charset=UTF-8");
        PrintWriter out =null;
        try {
            out = response.getWriter();

            List> list = new LinkedList>();
            if(channelId!=null && !channelId.equals("")){
                Channel tmp = new Channel();
                tmp.setChannelId(Integer.parseInt(channelId));
                subChannelList = subChannelDAO.listSubChannel(tmp);

                for(Object oo:subChannelList){
                    Map obj = new HashMap();
                    SubChannel c = (SubChannel)oo;
                    obj.put("subId", c.getSubId());
                    obj.put("subName", c.getSubName());
                    list.add(obj);
                }
            }
            String outStr = JsonUtil.toJSONStr(list);
            out.println(outStr);
        }catch(Exception e){
        e.printStackTrace();
        }
        out.flush();
        out.close();
        return null;
    }  

其实蛮简洁的
Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!