Question

假设我有对象数组，我需要使用之前的数据创建新的一个对象数组。在我的情况下，我需要过滤所有数据。 ES6的最佳解决方案是什么？

data = [{
 deviceId:1
 deviceStatus:"Offline"
 deviceType:"Tag"
 isConfigured:true
 isEnabled:true
 isLocalized:false
 isMasterAssigned:false
 lastAliveMessage:null
 location:{roomId: 5, name: "Room_207", zones: null, plan: null}
 name:"Tag For sending an alarm"
},{
 deviceId:2
 deviceStatus:"Online"
 deviceType:"Tag"
 isConfigured:true
 isEnabled:true
 isLocalized:false
 isMasterAssigned:false
 lastAliveMessage:null
 location:{roomId: 6, name: "Room_208", zones: null, plan: null}
 name:"Some Text"
}]

我只需要过滤所需的数据。

filteredData=[{
     deviceId:2
     deviceStatus:"Online"
     deviceType:"Tag"
     lastAliveMessage:null
     name: 'name:"Some Text'
     location: location.name
},{
     deviceId:2
     deviceStatus:"Online"
     deviceType:"Tag"
     lastAliveMessage:null
     name: 'name:"Some Text'
     location: location.name
}]

Answer 1

使用Array#map函数迭代它们并创建另一种类型的对象。

＆＃13;

const data = [{
    deviceId:1,
    deviceStatus:"Offline",
    deviceType:"Tag",
    isConfigured:true,
    isEnabled:true,
    isLocalized:false,
    isMasterAssigned:false,
    lastAliveMessage:null,
    location:{roomId: 5, name: "Room_207", zones: null, plan: null},
    name:"Tag For sending an alarm"
},{
    deviceId:2,
    deviceStatus:"Online",
    deviceType:"Tag",
    isConfigured:true,
    isEnabled:true,
    isLocalized:false,
    isMasterAssigned:false,
    lastAliveMessage:null,
    location:{roomId: 6, name: "Room_208", zones: null, plan: null},
    name:"Some Text"
}];

const newObj = data.map(item => ({
  deviceId: item.deviceId,
  deviceStatus: item.deviceStatus,
  deviceType: item.deviceType,
  lastAliveMessage: item.lastAliveMessage,
  name: item.name,
  location: item.location.name
}));

console.log(newObj);

＆＃13;

Answer 2

您可以为所需的密钥设置一个白名单。

> java.exe -jar myjar.jar MyClass """*.*"""

将对象数组缩减/过滤为新的对象数组。

2 个答案: