我正在尝试使用mongoose为我的mongodb数据库创建模型。这就是我想要做的事情:
var Class = mongoose.model('Class', {className: String, marks: [{type: Number}], grades: [{type: Number}]});
var User = mongoose.model('User', {email: String, classes: [Class] });
//Lets create a new user
var class1 = new Class({className: 'aaa', marks: [72, 88, 63], grades: [30, 40, 30]});
var user1 = new User({email: 'aaa@some.com', classes: [class1]});
保存class1
似乎没问题,但是当我查看mongodb时,会显示:{ "_id" : ObjectId("someId"), "className" : "TEST1234", "grades" : [ 30, 40, 30 ], "marks" : [ 72, 88, 63 ], "__v" : 0 }
什么是"__v : 0"
?
保存用户根本不成功,这是以下错误:ValidationError: CastError: Cast to Array failed for value "{ marks: [ 72, 88, 63 ],
grades: [ 30, 40, 30 ],
_id: someId,
className: 'TEST1234' }" at path "classes"
错误究竟是什么意思?为什么要向阵列投射任何东西?不应该classes: [Class]
是class
类型的数组吗?
答案 0 :(得分:73)
伙计,我有类似的问题创建这样的架构:
QuestionnaireSchema = mongoose.Schema({
formId: Number,
name: String,
questions: [
{
type: String,
title: String,
alternatives:[{
label: String,
value: "Mixed"
}]
}
]
});
我的错误是我正在使用"输入"作为字段名称,这是mongoose中的保留字。
我只是改变:
type: String,
到
formType: String,
并且有效。
答案 1 :(得分:9)
允许在名为type的属性上明确定义类型规则,并且不会抛出错误。像这样:
type: {type: String}
答案 2 :(得分:5)
尝试将类定义更改为:
public KafkaProducer(String zookeeperAddress, String topic) throws IOException,
KeeperException, InterruptedException {
this.zookeeperAddress = zookeeperAddress;
this.topic = topic;
ZooKeeper zk = new ZooKeeper(zookeeperAddress, 10000, null);
List<String> brokerList = new ArrayList<String>();
List<String> ids = zk.getChildren("/brokers/ids", false);
for (String id : ids) {
String brokerInfoString = new String(zk.getData("/brokers/ids/" + id, false, null));
Broker broker = Broker.createBroker(Integer.valueOf(id), brokerInfoString);
if (broker != null) {
brokerList.add(broker.connectionString());
}
}
props.put("serializer.class", KAFKA_STRING_ENCODER);
props.put("metadata.broker.list", String.join(",", brokerList));
producer = new Producer<String, String>(new ProducerConfig(props));
}
这是必需的,因为mongoose无法在没有相关模式的情况下解析对象。现在,当您为内部类对象创建一个新的Schema并在主userSchema中引用它时,mongoose应该能够解析您的对象。
答案 3 :(得分:4)
您的模型定义不正确,您应该修改如下。
// var Schema = mongoose.Schema;
var User = mongoose.model('User',{
email: String,
classes: [ {type: Schema.Types.ObjectID, ref: 'Class'}]
});
var Class1 = new Class({/*yourDataWillBeHere*/})
Class1.save(function(err, classData) {
var User1 = new User({/*YourDataWillBeHere*/})
User1.classes.push(classData._id);
User1.save(function(err, userData) {
//make something with userData object
})
})
然后,您可以使用populate()
这样的
User
.find()
.populate('classes')
.exec()
答案 4 :(得分:1)
现在,Mongoose支持子文档,这是嵌套数组的文档化方法,
If i < dgv.RowCount - 1 Then
<强>来源强>
答案 5 :(得分:1)
使用猫鼬5.7.0+和双嵌套模式时,我遇到了类似的问题。
除了与关键字type
无关,而是猫鼬验证错误。
https://github.com/Automattic/mongoose/issues/8472
临时解决方法:对子模式使用Schema.Types.Mixed
答案 6 :(得分:0)
默认情况下,如果模式中有一个键为'type'的对象,mongoose会将其解释为类型声明。
// Mongoose interprets this as 'loc is a String'
var schema = new Schema({ loc: { type: String, coordinates: [Number] } });
更改typeKey:
var schema = new Schema({
// Mongoose interpets this as 'loc is an object with 2 keys, type and coordinates'
loc: { type: String, coordinates: [Number] },
// Mongoose interprets this as 'name is a String'
name: { $type: String }
}, { typeKey: '$type' }); // A '$type' key means this object is a type declaration