Question

我有一个来自php文件的json输出，如下所示;

[{"device_id":"9700001","sensor_value":"31.5","update_time":"2017-04-28 18:49:06"},
{"device_id":"9700002","sensor_value":"31.5","update_time":"2017-04-28 18:47:05"},
{"device_id":"9700003","sensor_value":"31.5","update_time":"2017-04-28 18:45:05"},
{"device_id":"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:24:57"},
{"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:22:57"},
{"device_id":"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:20:56"},
{"device_id":"9700001","sensor_value":"33.1","update_time":"2017-04-28 06:18:56"},
{"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:16:56"},
{"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:14:56"}]

我想将此数组转换为特定数据集，即我想获取update_time的设备9700001的值和两个数据集的传感器值，对于9700001，

datasensor1=[{"sensor_value":"31.5","update_time":"2017-04-28 18:49:06"},{"sensor_value":"33.1","update_time":"2017-04-28 06:18:56"},{"sensor_value":"33.1","update_time":"2017-04-28 06:18:56"}]

用于设备9700002，

 datasensor2=[{"sensor_value":"31.5","update_time":"2017-04-28 18:47:05"},{"sensor_value":"33.1","update_time":"2017-04-28 06:22:57"},{"sensor_value":"33.1","update_time":"2017-04-28 06:16:56"},{"sensor_value":"33.1","update_time":"2017-04-28 06:14:56"}]

类似于设备9700003，

 datasensor3=[{"sensor_value":"31.5","update_time":"2017-04-28 18:45:05"},{"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:24:57"},{"sensor_value":"33.1","update_time":"2017-04-28 06:20:56"}]

我先厌倦了使用

按照device_id进行分组

 var sensor= _.groupBy(data,"device_id");

但我之后无法将对象分开。我该如何解决这个问题？

Answer 1

使用map()和var data = [{"device_id":"9700001","sensor_value":"31.5","update_time":"2017-04-28 18:49:06"}, {"device_id":"9700002","sensor_value":"31.5","update_time":"2017-04-28 18:47:05"}, {"device_id":"9700003","sensor_value":"31.5","update_time":"2017-04-28 18:45:05"}, {"device_id":"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:24:57"}, {"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:22:57"}, {"device_id":"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:20:56"}, {"device_id":"9700001","sensor_value":"33.1","update_time":"2017-04-28 06:18:56"}, {"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:16:56"}, {"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:14:56"}]; var sorted = {}; data.forEach((row) => { sorted[row.device_id] = []; }); Object.keys(sorted).forEach((id) => { sorted[id] = data.filter((row) => row.device_id === id).map((row) => ({ sensor_value: row.sensor_value, update_time: row.update_time })); }); console.log(sorted);：

＆＃13;

Action [cursor]· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

＆＃13;

Answer 2

尝试使用Array#reduce方法

＆＃13;

var data =[{"device_id":"9700001","sensor_value":"31.5","update_time":"2017-04-28 18:49:06"},
{"device_id":"9700002","sensor_value":"31.5","update_time":"2017-04-28 18:47:05"},
{"device_id":"9700003","sensor_value":"31.5","update_time":"2017-04-28 18:45:05"},
{"device_id":"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:24:57"},
{"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:22:57"},
{"device_id":"9700003","sensor_value":"33.1","update_time":"2017-04-28 06:20:56"},
{"device_id":"9700001","sensor_value":"33.1","update_time":"2017-04-28 06:18:56"},
{"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:16:56"},
{"device_id":"9700002","sensor_value":"33.1","update_time":"2017-04-28 06:14:56"}]

function group(data , id){
return data.reduce((a,b)=>{
if(b.device_id == id)
a.push({
sensor_value:b.sensor_value,
update_time:b.update_time
})
return a
},[])
}
console.log(group(data,9700001))
console.log(group(data,9700002))

＆＃13;

根据组将json数组简化为不同的变量

2 个答案: