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现在在公司用到了关于多级分类的菜单树，关于封装成树形结构传给前端，下面提供几个方式：
1：传统封装：每一级循环一次得到该子类
@Override public List<Map<String, Object>> getCatalist(Integer sch_id) { //得到一级导航 List<Map<String, Object>> list = courseDao.getCatalist(sch_id,0); //循环一级分类 for(Map<String, Object> map :list){ //获得一级分类的id值作为二级分类的父id Integer p_id = (Integer) map.get("p_code"); //得到二级分类 List<Map<String, Object>> list2 = courseDao.getCatalist(sch_id,p_id); if(list2 != null && list2.size()>0){ map.put("child", list2); } //循环二级分类 for(Map<String, Object> map2 :list2){ //获得二级分类的id值作为三级分类的父id Integer p_id2 = (Integer) map2.get("p_code"); //得到二级分类 List<Map<String, Object>> list3 = courseDao.getCatalist(sch_id,p_id2); if(list3 != null && list.size()>0){ map2.put("child", list3); } } } return list; }
2：递归封装
由于树形菜单的表示根据p_id来建立父子关系的，首先可以查询所有的菜单信息，不区分父子
@Override public List<CatagoryTree> getCatagory() { String sql = " SELECT cc.id,cc.p_id AS pid,cc.name AS cataname ,cc.sort_index , cc.status,cc.is_visable_nav nav FROM y_crs_catagory cc WHERE cc.sch_id=? ORDER BY sort_index"; List<Map<String, Object>> list = jdbcTemplateBaseDao.getSelectForList(sql, "9999"); List<CatagoryTree> cataDatas = new ArrayList<CatagoryTree>(); if(list!=null){ for(Map<String, Object> map : list){ CatagoryTree cataTree = new CatagoryTree(); cataTree.setId(String.valueOf(map.get("id"))); cataTree.setName(String.valueOf(map.get("cataname"))); cataTree.setPrientId(String.valueOf(map.get("pid"))); cataTree.setSortIndex(String.valueOf(map.get("sort_index"))); cataTree.setStatus(String.valueOf(map.get("status")).equals("true")?true:false); cataTree.setNav(String.valueOf(map.get("nav")).equals("true")?true:false); cataDatas.add(cataTree); } cataDatas=TreeUtils.treeCatagory(cataDatas, "0"); } return cataDatas; }
上面是将查询的所有分类信息封装成对象的集合cataDatas传入到分类树方法treeCatagory中做处理
public static List treeCatagory(List treeDataList, String parentID) { List list = null; CatagoryTree catagoryTree = null; if (treeDataList != null && treeDataList.size() > 0) { try { for (int i = 0; i < treeDataList.size(); i++) { catagoryTree = (CatagoryTree) treeDataList.get(i); if (catagoryTree.getPrientId().equals(parentID)) { catagoryTree.setChild(treeCatagory(treeDataList, catagoryTree.getId())); if (list == null) { list = new ArrayList(); } list.add(catagoryTree); } } } catch (Exception e) { // TODO: handle exception logger.error(catagoryTree.getName() + " 菜单，没有父节点"); e.printStackTrace(); } } return list; }
采用递归方式封装成树
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在这里写一个对list中的map集合进行排序的功能
sortField是map中需要排序的字段名
sortType是排序方式 asc或者desc
public static final List<Map<String, Object>> getSortedResult(List<Map<String, Object>> list, final Object sortField, final String sortType) { if(null != list && list.size() != 0 && null != sortField){ if("asc".equalsIgnoreCase(sortType) || "desc".equalsIgnoreCase(sortType)){ Collections.sort(list,new Comparator<Map<String,Object>>() { @Override public int compare(Map<String, Object> o1, Map<String, Object> o2) { if("asc".equalsIgnoreCase(sortType)) // 升序 return o1.get(sortField).toString().compareTo(o2.get(sortField).toString()); else // 降序 return o2.get(sortField).toString().compareTo(o1.get(sortField).toString()); } }); } } return list; } }
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原文地址：http://www.cnblogs.com/itliucheng/p/4663992.html
