node.js - 在 mongoose 结果上使用 lodash cloneDeep() - 无法更新生成的 JSON

Question

使用的技术:Node.js、Express、Mongoose、Mocha、lodash

在测试我的服务时，我的应用程序:

1. 使用 Mongoose 保存新文档获取结果
2. 使用 lodash.cloneDeep() 克隆结果
3. 尝试更改克隆实例的内容

结果是我得到了一个处于非常奇怪状态的克隆文档。当尝试修改值时，如果我查询已更改的特定对象的值，它会显示正确的值。但是，当我将文档本身转储到控制台时，就好像我没有更新这些值一样。同样，如果我将文档传递到保存例程中，它会将文档保存到 MongoDb，就好像我没有更改任何值一样。

这是代码:

测试代码:

describe('### UPDATE ALARM', function () {
    it('Should save an alarm record for newly created asset', async () => {
        var asset = await Asset.findOne({ _id: compAdminAsset })
        const req = { user: { _id: compAdminId } }
        var alarm = await AlarmController.getByAsset(req, asset)
        expect(alarm._user).to.be.equal(compAdminId)
        expect(alarm._asset).to.be.equal(compAdminAsset)

        var newAlarm = _.cloneDeep(alarm)
        console.log('#### Newly Cloned Alarm Object')
        console.log(newAlarm)

        // change the rhSettings values
        newAlarm.rhSettings.rhAlarmLow = true
        newAlarm.rhSettings.rhAlarmHigh = true
        newAlarm.rhSettings.rhLow = 40
        newAlarm.rhSettings.rhHigh = 85

        console.log()
        console.log('#### Cloned Alarm Object AFTER updating values')
        console.log(newAlarm)

        console.log()
        console.log('####  Directly query the values that were changed ')
        console.log('newAlarm.rhSettings.rhAlarmLow: ' + newAlarm.rhSettings.rhAlarmLow)
        console.log('newAlarm.rhSettings.rhAlarmHigh: ' + newAlarm.rhSettings.rhAlarmHigh)
        console.log('newAlarm.rhSettings.rhLow: ' + newAlarm.rhSettings.rhLow)
        console.log('newAlarm.rhSettings.rhLow:  '+ newAlarm.rhSettings.rhHigh)

        var savedAlarm = await AlarmController.update(req, newAlarm)
        .
        .
        .
    })
})

测试代码的输出:

    ### UPDATE ALARM
#### Newly Cloned Alarm Object
{ rhSettings:
   { rhLow: 0, rhHigh: 80, rhAlarmLow: false, rhAlarmHigh: false },
  tempSettings:
   { tempLow: 50,
     tempHigh: 100,
     tempAlarmLow: false,
     tempAlarmHigh: false },
  _user: 'c86a7618-2323-48ec-b9e0-0d953301e37f',
  _asset: '6ca7a1ba-fc16-4cbf-9b4d-029c508d6b6c',
  _id: 5bde26a5038bec31683b8d80,
  createdAt: 2018-11-03T22:52:21.604Z,
  updatedAt: 2018-11-03T22:52:21.604Z,
  __v: 0 }

#### Cloned Alarm Object AFTER updating values
{ rhSettings:
   { rhLow: 0, rhHigh: 80, rhAlarmLow: false, rhAlarmHigh: false },
  tempSettings:
   { tempLow: 50,
     tempHigh: 100,
     tempAlarmLow: false,
     tempAlarmHigh: false },
  _user: 'c86a7618-2323-48ec-b9e0-0d953301e37f',
  _asset: '6ca7a1ba-fc16-4cbf-9b4d-029c508d6b6c',
  _id: 5bde26a5038bec31683b8d80,
  createdAt: 2018-11-03T22:52:21.604Z,
  updatedAt: 2018-11-03T22:52:21.604Z,
  __v: 0 }

#### Query the values that were changed directly
newAlarm.rhSettings.rhAlarmLow: true
newAlarm.rhSettings.rhAlarmHigh: true
newAlarm.rhSettings.rhLow: 40
newAlarm.rhSettings.rhLow:  85

返回“newAlarm”对象的代码:

const Model = require('../models/alarm.model')

exports.getByAsset = async function (req, asset) {
    let newAlarm = new Model()
    newAlarm._asset = asset._id
    newAlarm._user = req.user._id
    newAlarm.rhSettings = asset.rhSettings
    newAlarm.tempSettings = asset.tempSettings
    return await newAlarm.save()
}

有什么想法导致克隆的 JSON 对象处于这种奇怪的状态吗？

Answer 1

为了让 Mongoose 对更改跟踪等感到满意，您将使用的方法是使用 set .

但是，我非常怀疑克隆对象并使用 set 是否与最佳实践相距甚远。

为什么不简单地创建另一个模型呢？为什么要通过克隆来规避 Mongoose 中的更改跟踪等构建？

您不妨简单地使用 toObject您的新 Mongoose 模型...跳过克隆...更改值等...然后只需将该对象传递给 new Model(yourObjectWithChanges) 然后保存:

var newAlarm = alarm.toObject() 
... do your changes
let newAlarmModel = new Model(newAlarm)
await newAlarmModel.save()