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  • Map与List数据操作

    为避免与数据库的多次连接,减少数据库的压力,先将所有的订货数据先从数据库中抽取出来,而后再将数据按门店进行分类汇总以备待用,Map与List混合操作,理解数据结构。提神醒脑哦。
    以下是原始数据结构:
    [{store_id=020, GoodsCode=04304654, Qty=8.000},
    {store_id=956, GoodsCode=04304685, Qty=7.000},
    {store_id=956, GoodsCode=04304715, Qty=6.000},
    {store_id=956, GoodsCode=04304722, Qty=2.000}]

    以下是想要的数据结构:
    [
    {store_id=020, orderList=[{GoodsCode=04304654, Qty=8.000}]},
    {store_id=956, orderList=[{GoodsCode=04304685, Qty=7.000},{GoodsCode=04304715, Qty=6.000},{GoodsCode=04304722, Qty=2.000}]}
    ]

    以下是实现代码:
    1.用一层循环从数据中找出门店的store_id;
    2.通过判断循环的store_id是否在新创建的数组中存在来保证其唯一性;
    3.在用唯一的store_id来获取属于该门店的订货信息

     1 public List<Map<String, Object>> allOrderedStore(List<Map<String, Object>> allStoreOrderDetails) {
     2 List<String> storeList = new ArrayList<String>();
     3 List<Map<String, Object>> allOrderedStoreList =new ArrayList<Map<String, Object>>();
     4 
     5 for (int i = 0; i < allStoreOrderDetails.size(); i++) 
     6 { 
     7 String storeId=allStoreOrderDetails.get(i).get("store_id").toString();
     8 
     9 if (!storeList.contains(storeId)) 
    10 { 
    11 storeList.add(storeId);
    12 Map<String, Object> Map = new HashMap<String, Object>();
    13 Map.put("storeid", storeId);
    14 List<Map<String, Object>> orderlist =new ArrayList<Map<String, Object>>();
    15 for (int j = 0; j < allStoreOrderDetails.size(); j++) {
    16 if(storeId.equals(allStoreOrderDetails.get(j).get("store_id"))){
    17 Map<String, Object> orderMap = new HashMap<String, Object>();
    18 orderMap.put("GoodsCode", allStoreOrderDetails.get(j).get("GoodsCode"));
    19 orderMap.put("Qty", allStoreOrderDetails.get(j).get("Qty"));
    20 orderlist.add(orderMap);
    21 }
    22 }
    23 Map.put("orderList", orderlist);
    24 allOrderedStoreList.add(Map);
    25 }
    26 }
    27 return allOrderedStoreList;
    28 }
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  • 原文地址:https://www.cnblogs.com/maoBable/p/8964483.html
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