json:java
{"updated_at":1551780617,"attr":{"uptime_h":3,"uptime_m":17},"did":"GBBxjJYAxE4apkxwEzR3"}json
Map rMap = (Map) JSON.parse("{"updated_at":1551780617,"attr":{"uptime_h":3,"uptime_m":17},"did":"GBBxjJYAxE4apkxwEzR3"} ");spa
在提取出 rMap.get("attr"); 時獲得的是:{"uptime_h":3,"uptime_m":17}對象
但我要獲得裏面的東西,直接用字符串很差搞,我把它轉爲json 提取:字符串
Map rMap2 = (Map) JSON.parse((String)rMap.get("attr"));get
結果就報錯了: com.alibaba.fastjson.JSONObject cannot be cast to java.lang.Stringast
緣由是:改map提取出來的對象不能轉爲String,而要經過它的方法 toString 來轉化:date
String str = map.get("attr").toString();
System.out.println("ddd===" + str);
Map rMap = (Map) JSON.parse(str);
System.out.println(rMap.get("uptime_h"));map
OK!!!方法