mongodb中存储数组和新数组的区别

发布于 2024-11-05 02:46:17 字数 502 浏览 0 评论 0原文

我是 mongodb 的新手。

我在 mongodb 中存储了以下数据

    { "_id" : ObjectId("4"), "uid" : "1", "friendsIds" : [ "2", "3" ] }

情况 1： 现在，我有一组新的 FriendsIds ["2","3","4"]

我想比较这两个集合，并获取未存储在数据库中的值。在本例中，我想要得到 4

情况 2： 现在，我有一组新的 FriendsIds ["3","4"]

我想比较这两个集合，并获取存储在数据库中但不存储在新集合中的值。在这种情况下，我想得到2

情况2可以使用mapReduce完成吗？

mongodb中这两种情况如何实现？

原文

I am new to mongodb.

I have the following data stored in mongodb

    { "_id" : ObjectId("4"), "uid" : "1", "friendsIds" : [ "2", "3" ] }

Case 1:
Now, I have new set of friendsIds ["2","3","4"]

I want to compare the both the sets, and get the value which is not stored in database. In this case, I want to get 4

Case 2:
Now, I have new set of friendsIds ["3","4"]

I want to compare the both the sets, and get the value which is stored in database but not in the new set. In this case, I want to get 2

Can the Case 2 be done using mapReduce?

How to achieve both the cases in mongodb?

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谈场末日恋爱 2024-11-12 02:46:17

对于情况 1，您似乎需要 [$addToSet][1]。看一下 $each 示例。

对于情况2，没有这样的服务器端功能。只需带回文档并在客户端进行比较即可。

您正在做什么操作，您只想要 2 而不是其他一切？

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攒眉千度 2024-11-12 02:46:17

如果 ruby 中有两个数组

a = [ "2", "3" ]
b = ["2","3","4"]

：

a-b = ["4"]

if you have two arrays

a = [ "2", "3" ]
b = ["2","3","4"]

in ruby :

a-b = ["4"]

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萤火眠眠 2024-11-12 02:46:17

这可以在聚合框架中使用其集合运算符来完成:

> db.foo.insert({"friendsIds" : [ "2", "3"]})
WriteResult({ "nInserted" : 1 })

> db.foo.aggregate([ {$project: {diff: {$setDifference: [["2","3","4"], '$friendsIds']}}} ])
{ "_id" : ObjectId("53490a53a0e3abbe0dd0c21c"), "diff" : [ "4" ] }

> db.foo.aggregate([ {$project: {diff: {$setDifference: ['$friendsIds', ["3","4"]]}}} ])
{ "_id" : ObjectId("53490a53a0e3abbe0dd0c21c"), "diff" : [ "2" ] }

This can be done in the aggregation framework using its set operators:

> db.foo.insert({"friendsIds" : [ "2", "3"]})
WriteResult({ "nInserted" : 1 })

> db.foo.aggregate([ {$project: {diff: {$setDifference: [["2","3","4"], '$friendsIds']}}} ])
{ "_id" : ObjectId("53490a53a0e3abbe0dd0c21c"), "diff" : [ "4" ] }

> db.foo.aggregate([ {$project: {diff: {$setDifference: ['$friendsIds', ["3","4"]]}}} ])
{ "_id" : ObjectId("53490a53a0e3abbe0dd0c21c"), "diff" : [ "2" ] }

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