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      1. 用于 Web 照片庫的正確 NoSQL 數據架構

        Proper NoSQL data schema for web photo gallery(用于 Web 照片庫的正確 NoSQL 數據架構)

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                  本文介紹了用于 Web 照片庫的正確 NoSQL 數據架構的處理方法,對大家解決問題具有一定的參考價值,需要的朋友們下面隨著小編來一起學習吧!

                  問題描述

                  我正在尋找為照片庫的 NoSQL 存儲構建合適的數據結構.在我的 Web 應用程序中,一張照片可以是 1 個或多個相冊的一部分.我有使用 MySQL 的經驗,但幾乎沒有使用過鍵值存儲.

                  I'm looking to build an appropriate data structure for NoSQL storage of a photo gallery. In my web application, a photo can be part of 1 or more albums. I have experience with MySQL, but almost none with key-value storage.

                  使用 MySQL,我將設置 (3) 個表,如下所示:

                  With MySQL, I would have set up (3) tables as follows:

                  photos (photo_id, title, date_uploaded, filename)
                  albums (album_id, title, photo_id)
                  album_photo_map (photo_id, album_id)
                  

                  然后,要檢索 5 張最新照片的列表(帶有相冊數據),查詢如下:

                  And then, to retrieve a list of the 5 latest photos (with album data), a query like this:

                  SELECT *
                  FROM albums, photos, album_photo_map
                  WHERE albums.album_id = album_photo_map.album_id AND
                                  photos.photo_id = album_photo_map.photo_id
                  ORDER BY photos.date_uploaded DESC LIMIT 5;
                  

                  如何使用 NoSQL 鍵值對數據庫完成類似的查詢?(特別是亞馬遜的 DynamoDB.)存儲會是什么樣子?索引如何工作?

                  How would I accomplish a similar query using a NoSQL key-value pair database? (Specifically, Amazon's DynamoDB.) What would the storage look like? How would the indexing work?

                  推薦答案

                  使用 mongodb lingo,您的集合可能如下所示:

                  Using mongodb lingo, your collections could look like this:

                  photos = [
                      {
                          _id: ObjectId(...),
                          title: "...",
                          date_uploaded: Date(...),
                          albums: [
                              ObjectId(...),
                              ...
                          ]
                      },
                      ...
                  ]
                  
                  albums = [
                      {
                          _id: ObjectId(...),
                          title: "..."
                      }
                  ]
                  

                  查找 5 張最新照片的方法如下:

                  Finding the 5 newest photos would be done like this:

                  > var latest = db.photos.find({}).sort({date_uploaded:1}).limit(5);
                  

                  mongo 中沒有服務器端連接,因此您必須像這樣獲取所有最新專輯:

                  There's no server-side joins in mongo, so you'd have to fetch all the latest albums like this:

                  > var latest_albums = latest.find({}, {albums: 1});
                  

                  當然,那你必須把它歸結為一個集合.

                  Of course, then you have to boil this down into a set.

                  如果您只是將相冊嵌入照片文檔中,實際上會更容易,因為它們很小:

                  It's actually easier if you just embed the album inside the photo documents, since they're small:

                  photos = [
                      {
                          _id: ObjectId(...),
                          title: "...",
                          date_uploaded: Date(...),
                          albums: [
                              {name: "family-vacation-2011", title: "My family vacation in 2010"},
                              ...
                          ]
                      },
                      ...
                  ]
                  

                  那么查詢也是一樣的,只是你不必加入.查找相冊中的所有照片如下所示:

                  Then querying is the same, but you don't have to join. Finding all photos in an album looks like:

                  > db.photos.find({albums:{$elemMatch:{name: "family-vacation-2011"}}});
                  

                  這篇關于用于 Web 照片庫的正確 NoSQL 數據架構的文章就介紹到這了,希望我們推薦的答案對大家有所幫助,也希望大家多多支持html5模板網!

                  【網站聲明】本站部分內容來源于互聯網,旨在幫助大家更快的解決問題,如果有圖片或者內容侵犯了您的權益,請聯系我們刪除處理,感謝您的支持!

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