Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
416 views
in Technique[技术] by (71.8m points)

aggregation framework - What's the $unwind operator in MongoDB?

This is my first day with MongoDB so please go easy with me :)

I can't understand the $unwind operator, maybe because English is not my native language.

db.article.aggregate(
    { $project : {
        author : 1 ,
        title : 1 ,
        tags : 1
    }},
    { $unwind : "$tags" }
);

The project operator is something I can understand, I suppose (it's like SELECT, isn't it?). But then, $unwind (citing) returns one document for every member of the unwound array within every source document.

Is this like a JOIN? If yes, how the result of $project (with _id, author, title and tags fields) can be compared with the tags array?

NOTE: I've taken the example from MongoDB website, I don't know the structure of tags array. I think it's a simple array of tag names.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

First off, welcome to MongoDB!

The thing to remember is that MongoDB employs an "NoSQL" approach to data storage, so perish the thoughts of selects, joins, etc. from your mind. The way that it stores your data is in the form of documents and collections, which allows for a dynamic means of adding and obtaining the data from your storage locations.

That being said, in order to understand the concept behind the $unwind parameter, you first must understand what the use case that you are trying to quote is saying. The example document from mongodb.org is as follows:

{
 title : "this is my title" ,
 author : "bob" ,
 posted : new Date () ,
 pageViews : 5 ,
 tags : [ "fun" , "good" , "fun" ] ,
 comments : [
             { author :"joe" , text : "this is cool" } ,
             { author :"sam" , text : "this is bad" }
 ],
 other : { foo : 5 }
}

Notice how tags is actually an array of 3 items, in this case being "fun", "good" and "fun".

What $unwind does is allow you to peel off a document for each element and returns that resulting document. To think of this in a classical approach, it would be the equivilent of "for each item in the tags array, return a document with only that item".

Thus, the result of running the following:

db.article.aggregate(
    { $project : {
        author : 1 ,
        title : 1 ,
        tags : 1
    }},
    { $unwind : "$tags" }
);

would return the following documents:

{
     "result" : [
             {
                     "_id" : ObjectId("4e6e4ef557b77501a49233f6"),
                     "title" : "this is my title",
                     "author" : "bob",
                     "tags" : "fun"
             },
             {
                     "_id" : ObjectId("4e6e4ef557b77501a49233f6"),
                     "title" : "this is my title",
                     "author" : "bob",
                     "tags" : "good"
             },
             {
                     "_id" : ObjectId("4e6e4ef557b77501a49233f6"),
                     "title" : "this is my title",
                     "author" : "bob",
                     "tags" : "fun"
             }
     ],
     "OK" : 1
}

Notice that the only thing changing in the result array is what is being returned in the tags value. If you need an additional reference on how this works, I've included a link here. Hopefully this helps, and good luck with your foray into one of the best NoSQL systems that I have come across thus far.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

56.9k users

...