MongoDB使用外键将字段复制到另一个集合

时间:2013-04-06 02:32:22

标签: mongodb copy foreign-keys database

我想将用户集中的颜色复制到Car集合中。我使用的是外键,即userID。

> db.test1.User.find()
{ "_id" : ObjectId("515f7db83f71d6bcb1c41a48"), "age" : 33, "Color" : "blue" }
{ "_id" : ObjectId("515f7dc03f71d6bcb1c41a49"), "age" : 52, "Color" : "red" }
{ "_id" : ObjectId("515f7dc43f71d6bcb1c41a4a"), "age" : 43, "Color" : "yellow" }


> db.test2.Car.find()
{ "_id" : ObjectId("515f84883f71d6bcb1c41a54"), "speed" : 291, "userID" : ObjectId("515f7db83f71d6bcb1c41a48") }
{ "_id" : ObjectId("515f84883f71d6bcb1c41a55"), "speed" : 202, "userID" : ObjectId("515f7db83f71d6bcb1c41a49") }
{ "_id" : ObjectId("515f84883f71d6bcb1c41a56"), "speed" : 193, "userID" : ObjectId("515f7db83f71d6bcb1c41a4a") }

这是我的查询

db.test1.User.find().forEach( 
function(x)
{
  db.test2.Car.update( { userID: x._id }, { $set: { color: x.color} } ) 
} );

我想要这个结果:

> db.test2.Car.find()
{ "_id" : ObjectId("515f84883f71d6bcb1c41a54"), "speed" : 291, "userID" : ObjectId("515f7db83f71d6bcb1c41a48"), "color" : "blue" }
{ "_id" : ObjectId("515f84883f71d6bcb1c41a55"), "speed" : 202, "userID" : ObjectId("515f7db83f71d6bcb1c41a49"), "color" : "red" }
{ "_id" : ObjectId("515f84883f71d6bcb1c41a56"), "speed" : 193, "userID" : ObjectId("515f7db83f71d6bcb1c41a4a"), "color" : "yellow" }

感谢您的帮助!

2 个答案:

答案 0 :(得分:4)

您的测试设置存在几个问题:

  • 字段名称不匹配(复制时引用color而不是Color
  • 目标集合中只有一个示例外键匹配:ObjectId('515f7db83f71d6bcb1c41a48')
  • 您的更新只会影响“外键”的第一个匹配文档。对于1:1的关系,这是好的,但不是1:多

将上述内容考虑在内的更正示例(除非匹配键外):

db.test1.User.find().forEach( 
    function(x) {
        db.test2.Car.update(
            // query 
            { userID: x._id },

            // update 
            { $set: { color: x.Color} },

            // options:
            { "multi" : true } // Update all matching documents
        );
    }
);

这导致为示例文档中实际匹配的唯一外键设置{color:blue}

db.test2.Car.find()
{
    "_id" : ObjectId("515f84883f71d6bcb1c41a55"),
    "speed" : 202,
    "userID" : ObjectId("515f7db83f71d6bcb1c41a49")
}
{
    "_id" : ObjectId("515f84883f71d6bcb1c41a56"),
    "speed" : 193,
    "userID" : ObjectId("515f7db83f71d6bcb1c41a4a")
}
{
    "_id" : ObjectId("515f84883f71d6bcb1c41a54"),
    "color" : "blue",
    "speed" : 291,
    "userID" : ObjectId("515f7db83f71d6bcb1c41a48")
}

答案 1 :(得分:0)

这是我的两分钱:您可以放下“ function(x)”。我更喜欢以下语法:

db.source_collection_name.find().forEach(
  doc=>
      db.target_collection_name.update(
        {"_id": doc._id},
        {$set: {"field_to_update": doc.field}}
      )
)

在您的情况下,类似这样的方法应该起作用:

db.test1.User.find().forEach(
  doc=> 
    db.test2.Car.update( 
      { "userID": doc._id }, 
      { $set: { "color": doc.Color} } 
    ) 
);