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 !!!-----------