NoSQL、DBRefs：MongoDB

Question

如何自己根据给定字段创建DBRef？例如，我在客户表中有 UID 字段，在订单表中也有这个 UID 字段，它将用于引用此请求所属的客户，它只是工作得不太好，请参阅：

> db.customers.findOne();
{
        "_id" : ObjectId("4ef4a61a90eec3e3c748263c"),
        "uid" : 1,
        "name" : "Andrey",
        "lastname" : "Knupp Vital"
}
> db.orders.findOne();
{
        "_id" : ObjectId("4ef4a66490eec3e3c748263d"),
        "oid" : 1,
        "uid" : 1,
        "price" : "149.90"
}
> db.orders.remove();
> order = { oid : 1 , price : 149.90 , uid : new DBRef ( 'customers' , 1 ) } ;
{
        "oid" : 1,
        "price" : 149.9,
        "uid" : {
                "$ref" : "customers",
                "$id" : 1
        }
}
> db.orders.save ( order ) ;
> order.uid.fetch();
null
> order.uid
{ "$ref" : "customers", "$id" : 1 }
>

Answer 1

DBRef $id 值必须始终设置为引用文档的 _id 字段的值。在你的例子中你没有这样做。固定版本：

> db.customers.findOne()
{
        "_id" : ObjectId("4ef4a61a90eec3e3c748263c"),
        "uid" : 1,
        "name" : "Andrey",
        "lastname" : "Knupp Vital"
}
> db.orders.findOne()
{
        "_id" : ObjectId("4ef4a66490eec3e3c748263d"),
        "oid" : 1,
        "uid" : 1,
        "price" : "149.90"
}
> db.orders.remove()
> order = { oid : 1 , price : 149.90 , uid : new DBRef ( 'customers' , ObjectId("4ef4a61a90eec3e3c748263c") ) } ;
{
        "oid" : 1,
        "price" : 149.9,
        "uid" : {
                "$ref" : "customers",
                "$id" : ObjectId("4ef4a61a90eec3e3c748263c")
        }
}
> db.orders.save(order)
> order.uid.fetch()
{
        "_id" : ObjectId("4ef4a61a90eec3e3c748263c"),
        "uid" : 1,
        "name" : "Andrey",
        "lastname" : "Knupp Vital"
}

Answer 2

我手动做。拥有额外的标识符字段可能最终会导致出现问题。在您的订单文档中，创建“customer_id”字段并将其设置为客户的 _id。这就像使用关系引用创建 SQL 外键。我在 PHP 中这样做：

$customer = $this->mongo->db->customer->findOne(array("name" => $customer_name));
$customer_id = new MongoID($customer['_id']);
$order = Array(
            'customer_id' => $customer_id,
        ...
        );

$this->mongo->db->order->insert($order);