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  • 关于多级分类的封装

    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

    ------------------------------------------------------------------------------------------------------------------------------

    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:https://www.cnblogs.com/itliucheng/p/4663992.html
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