2018_01_17 处理复杂的JSON格式数据

Map<String,List<RoadOperationVehicle>> map = JSON.parseObject(tempString,new TypeReference<Map<String,List<RoadOperationVehicle>>>(){});

使用fastJson处理复杂的json，json格式为：

{"data":[{"linkid":"49793713","trans":11,"speed_num":1,"tired_num":1,"total_num":3},{"linkid":"58899509","trans":30,"speed_num":1,"tired_num":1,"total_num":1}........