11

我正在运行猫鼬快速入门fluffy.speak(),我的应用程序一直因错误而死TypeError: Object { name: 'fluffy', _id: 509f3377cff8cf6027000002 } has no method 'speak'

本教程中我的(稍作修改)代码:

"use strict";

var mongoose = require('mongoose')
  , db = mongoose.createConnection('localhost', 'test');

db.on('error', console.error.bind(console, 'connection error:'));
db.once('open', function () {
    var kittySchema = new mongoose.Schema({
        name: String
    });
    var Kitten = db.model('Kitten', kittySchema);
    var silence = new Kitten({name: 'Silence'});
    console.log(silence.name);
    kittySchema.methods.speak = function() {
        var greeting = this.name ? "Meow name is" + this.name : "I don't have a name";
        console.log(greeting);
    };

    var fluffy = new Kitten({name: 'fluffy'});

    fluffy.speak();

    fluffy.save(function(err) {
        console.log('meow');
    });

    function logResult(err, result) {
        console.log(result);
    }

    Kitten.find(logResult);
    Kitten.find({name: /fluff/i }, logResult);
});
4

1 回答 1

9

当您调用db.model时,模型是从您的模式编译的。那时,它schema.methods被添加到模型的原型中。因此,您需要先在模式上定义任何方法,然后再用制作模型。

// ensure this method is defined before...
kittySchema.methods.speak = function() {
    var greeting = this.name ? "Meow name is" + this.name : "I don't have a name";
    console.log(greeting);
}

// ... this line.
var Kitten = db.model('Kitten', kittySchema);

// methods added to the schema *afterwards* will not be added to the model's prototype
kittySchema.methods.bark = function() {
    console.log("Woof Woof");
};

(new Kitten()).bark(); // Error!  Kittens don't bark.
于 2012-11-11T05:22:38.497 回答