Question

var mongoose = require('mongoose')
mongoose.connect('mongodb://127.0.0.1/DocTest');

var patientsSchema = mongoose.Schema({
//This is the value I wanna populate in the rdvs collection.

ssn: String
//But I can't see to have it working.

});


var patients = mongoose.model('patients', patientsSchema);

var Max = new patients({
ssn: "okokok" 
});
Max.save(function (err) {
if (err) {
    console.log(err);
}
else {
    console.log('wink')
}
});

var rdvsSchema = mongoose.Schema({
Heure: {
    type: Date
},
patient: {
    type: mongoose.Schema.Types.ObjectId, ref: 'patients'
}
});

var rdvs = mongoose.model('rdvs', rdvsSchema);

var rdvs1 = new rdvs({
Heure: 14
}
);
rdvs1.save(function (err) {
if (err) {
    console.log(err);
}
else {
    console.log('wonk')
}
});

这是我正在尝试工作的请求：

rdvs.findOne().populate('patient').exec(function(err, rdvs){
console.log (rdvs.patient.ssn)
});

我在这里苦苦挣扎，这里的问题是我想将患者的ssn值添加到我的rdvs集合中。

Answer 1

Mongoose操作是asynchronous。在执行其他操作之前，您需要等待操作完成。在您的情况下，第二个查询取决于您的第一个查询的响应 - 因此您需要嵌套您的操作。

// setup schemas
var patientSchema = mongoose.Schema({
    ssn: String
});

var rdvSchema = mongoose.Schema({
    Heure: Number, // changed: you're expecting a number
    patient: { type: mongoose.Schema.Types.ObjectId, ref: 'Patient' }
});

var Patient = mongoose.model('Patient', patientSchema);
var Rdv = mongoose.model('Rdv', rdvSchema);

// create a new patient
Patient.create({ ssn: "okokok" }, function (err, patient) {
    if (err) {
        console.log(err);
        return;
    }
    if (!patient) {
        console.log('patient could not be created');
        return;
    }
    // you can only create a RDV once a patient has been created
    Rdv.create({ Heure: 14, patient: patient._id }, function (err, rdv) {
        if (err) {
            console.log(err);
            return;
        }
        if (!rdv) {
            console.log('rdv could not be created');
            return;
        }
        // send response
    });
});

除此之外：养成正确命名变量的习惯。通常，models是单数且大写的，例如Patient和documents是小写的，例如patient。

.populate（）引用问题

1 个答案: