如何在CouchDB中编写SELECT COUNT（DISTINCT字段）查询？ [英] How do you write a SELECT COUNT(DISTINCT field) query in CouchDB?
问题描述
在CouchDB中是否有一种很好的方法来模仿SELECT COUNT（DISTINCT字段）的行为？
Is there a good way to mimic the behavior of SELECT COUNT(DISTINCT field) in CouchDB?
想象一下我们有以下文档，它记录了时间用户播放了某首歌曲：
Imagine we have the following document, which records the time at which a user played a certain song:
{
song_id: "happy birthday",
user_id: "boris",
date_played: [2011, 11, 14, 00, 12, 55],
_id: ...
}
我想知道用户 boris曾经播放过的不同歌曲的数量。如果我们的用户已经听了20次生日快乐，那首歌仍然只占总歌曲数的+1。
I'd like to know the number of distinct songs ever played by our user "boris". If our user has listened to "happy birthday" 20 times, that song should still contribute just +1 to the overall song count.
在MySQL中，我只需执行 SELECT COUNT（DISTINCT song_id）FROM播放WHERE user_id = boris ，但是在CouchDB中编写此代码时我空白。
In MySQL, I'd simply execute SELECT COUNT(DISTINCT song_id) FROM plays WHERE user_id = "boris", but I'm drawing a blank when it comes to writing this in CouchDB.
工作环境1：如果我更改了架构，而是将所有歌曲播放存储在单个用户文档中的 boris，那么我可以写一个映射到仅发出不同的值。但是，如果我想在last.fm的规模上构建某些东西，我担心的是，随着 boris文档大小（播放次数）的持续增长，更新将花费很长时间。 （可能还会有一个最终要达到的最大文档大小。）
Work-Around 1: If I changed my schema and instead stored all the song-plays inside a single user document for "boris" I could then write a map to emit only distinct values. However, if I wanted to build something on the scale of last.fm, my fear is that updates would start taking a very long time as the "boris" document size (number of plays) continued to grow. (There might also be a maximum document size that I would eventually hit).
工作环境2：我也可以编写一个map函数返回 all 的不同记录，我的python脚本可以总结这些记录；但是再加上成千上万首不同的歌曲，这也会变得非常慢。
Work-Around 2: I could also write a map function to return all of the distinct records, which my python script could sum up itself; but again with hundreds of thousands of distinct songs this would become very slow as well.
我还缺少其他选择吗？
推荐答案
此答案由Zachary Zolton在沙发上的邮件列表中提供：
This answer was provided by Zachary Zolton on the couchdb mailing list:
开始您已经有了一个视图，可以为您提供鲍里斯（Boris）的5万首独特的
歌曲，您可以使用_list函数返回行数。
Since you've already got a view that'll give you Boris's 50k unique songs, you could use a _list function to return the number of rows.
像这样应该可以解决问题：
Something like this should do the trick:
function() {
var count = 0;
while(getRow()) count++;
return JSON.stringify({count: count});
}
如果使用相同的视图，键范围和$查询此列表函数b $ b组级别，它将仅使用一些JSON进行响应，例如： { count： 50612}
If you query this list function, with the same view, key range and
group level, it'll just respond with a bit of JSON, such as: {"count":"50612"}
您可以在此处阅读更多内容：
You can read up more here:
- http://guide.couchdb.org/draft/transforming.html
- http://wiki.apache.org/couchdb/Formatting_with_Show_and_List
- http://guide.couchdb.org/draft/transforming.html
- http://wiki.apache.org/couchdb/Formatting_with_Show_and_List
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