我已经重写了我的代码,现在我有三个灯,我一个接一个地关闭。有没有办法将这三个变量都放在一个变量中,只运行一次关闭命令?
// Turn off a bunch of lights
// Specify ID for a specific lamp
let KitchentableRight = await Homey.devices.getDevice({id: 'ef6bae9c-ec57-46ff-9011-905f90d16589'});
let KitchentableMiddle = await Homey.devices.getDevice({id: 'e29ab963-5139-4fdb-b4ec-ec25c103a62a'});
let KitchentableLeft = await Homey.devices.getDevice({id: '8b102bc6-00dd-4305-b111-f67a53ffd9d9'});
// Turn off tha lights based on their ID:s
KitchentableRight.setCapabilityValue('onoff', false);
KitchentableMiddle.setCapabilityValue('onoff', false);
KitchentableLeft.setCapabilityValue('onoff', false);
return true发布于 2018-10-17 00:03:09
要使循环对每个设备运行,它应该是:
devices.forEach(device => {
if(device.name) return false;
console.log(device.name);
device.setCapabilityValue('onoff', !device.state.onoff);
})如果a是设备名称的数组,则可以通过:a.indexOf(device.name) != -1检查该设备是否在
https://stackoverflow.com/questions/52839387
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