猫鼬TypeError:用户不是构造函数

问题描述

我正在尝试使用Mongoose和MongoDB将子文档添加到父架构中，但是我被抛出以下错误:

I'm trying to add a subdocument to a parent schema with Mongoose and MongoDB however I'm being thrown the following error:

TypeError: User is not a constructor

这基于Monaose在子文档上的文档，我认为所有内容都是相同的.我该如何进一步调试呢?

This is based off Mongoose's documentation on subdocuments and I think everything is the same. How can I debug this further?

路由器

// Add a destination to the DB
router.post('/add', function(req, res, next) {
  let airport = req.body.destination
  let month = req.body.month
  let id = (req.user.id)

  User.findById(id , function (err, User) {
    if (err) return handleError(err)

    function addToCart (airport, month, id) {
      var user = new User ({
        destinations: [(
          airport = '',
          month = ''
        )]
      })

      dog.destinations[0].airport = airport
      dog.destinations[0].month = month
      dog.save(callback)
      res.status(200).send('added')
    }
    addToCart()
  })
  console.log(airport)
})

模式

var destinationSchema = new Schema({
  airport: String,
  month: String
})

// Define the scheme
var User = new Schema ({
  firstName: {
    type: String,
    index: true
  },
  lastName: {
    type: String,
    index: true
  },
  email: {
    type: String,
    index: true
  },
  homeAirport: {
    type: String,
    index: true
  },
  destinations: [destinationSchema]
})


User.plugin(passportLocalMongoose)

module.exports = mongoose.model('User', User)

推荐答案

JavaScript对变量名区分大小写.您有User模型和User结果具有相同的名称.

JavaScript is case sensitive about the variable names. You have User model and the User result with the same name.

您的代码将进行以下更改:

Your code will work with the following change :

   User.findById(id , function (err, user) {
/*                                   ^ use small `u` */
       if (err) return handleError(err)

/* rest of your code */

还请记住，在代码中进一步声明了另一个名为user的变量.您需要将其更改为其他内容.

Also keep in mind that further in your code you are declaring another variable named user. You will need to change that to something different.

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