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!
答案 0 :(得分:4)
您的测试设置存在几个问题:
color
而不是Color
)ObjectId('515f7db83f71d6bcb1c41a48')
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} }
)
);