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  • JSON 新感

    1,如何解析复杂JSON 

          选择fastJson 用起来简单方便,前提必须是标准的JSON;

    例子:

    paraArray = "[" + paraArray + "]";
                    _finalMap.put("sourcename", name);
                    _finalMap.put("tablename", tablename);
                    _finalMap.put("datasource_id", datasource_id);
                    _finalMap.put("id", id);
                }
                list = JSON.parseArray(paraArray, Map.class);
                if(list==null){
                    return "{"code":"444","msg":"查询的信息不存在"}";
                }
                // 后续自己处理即可
                Map map1 = list.get(0);
                List list1 = (List) map1.get("tableInfos");
                Object ob = null;
                for (int i = 0; i < list1.size(); i++) {
                    ob = (Object) list1.get(i);
                
                String sob = ob.toString();
                sob = "[" + sob + "]";
                List list3 = JSON.parseArray(sob, Map.class);
                System.out.println(list3.toString());
                Map map3 = (Map) list3.get(0);
                List list4 = (List) map3.get("fieldSelectArray");
                List fieldSelectArrayList = new ArrayList();
                for (int k = 0; k < list4.size(); k++) {
                    Map temp = new HashMap();
                    Object o = list4.get(k);
                    String so = o.toString();
                    so = "[" + so + "]";
                    List list5 = JSON.parseArray(so, Map.class);
                    System.out.println(list5);
                    Map map4 = (Map) list5.get(0);
                    String _name = (String) map4.get("name");
                    String _type = (String) map4.get("type");
                    String _dict = (String) map4.get("dict");
                    temp.put("name", _name);
                    temp.put("type", _type);
                    temp.put("dict", _dict);
                    fieldSelectArrayList.add(temp);

    通过获取的字符串加入 “[]”,转化为List<map> ,获取map值,然后再添加[] 再次转化,最终得到,想要的

    2,-------字符串评价为JSON

    注意:大小包含的JSONObject jsons = new JSONObject();

                  jsons.put("paramname",paramname);
                jsons.put("type", type);
                jsons.put("dict",dict);
                  
                JSONObject jsons1 = new JSONObject();
                jsons1.put("paramname", "A");
                jsons1.put("type","B");
                jsons1.put("dict","C");
                
                List<JSONObject> st = new ArrayList<JSONObject>();
                st.add(jsons);
                st.add(jsons1);
                
                  JSONObject json = new JSONObject();
                  json.put("sourcename", name);
                  json.put("tableName", tablename);
                  json.put("datasource_id", datasource_id);
                  json.put("id", id);
                  json.put("fieldSelectArray", st);
                  
                  
                  JSONObject jsonB = new JSONObject();
                  jsonB.put("tableInfos", json);
                  
         
    System.out.println(jsonB.toString());


    先添加小的,最内部的JSOn,然后依次向外添加,----------得到想要的为止。


    ------------GOOD LUCKLY !!!-----------
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  • 原文地址:https://www.cnblogs.com/xiaohu1218/p/8098243.html
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