MongoDB独特的聚合

本文介绍了MongoDB独特的聚合的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在研究一个查询，以查找每个州的邮政编码最多的城市:

I'm working on a query to find cities with most zips for each state:

db.zips.distinct("state", db.zips.aggregate([ {$group:{_id:{state:"$state", city:"$city"},numberOfzipcodes:{$sum:1}}}, {$sort:{numberOfzipcodes:-1}}]))

查询的聚合部分似乎可以正常工作，但是当我添加非重复项时，我得到的结果为空.

The aggregate part of the query seems to work fine, but when I add the distinct I get an empty result.

这是因为我的ID中有州吗?我可以做类似distinct("_id.state的事情吗?

Is this because I have state in the id? Can I do something like distinct("_id.state ?

推荐答案

不同的聚合框架不可互操作.

Distinct and the aggregation framework are not inter-operable.

相反，您只想要:

db.zips.aggregate([ {$group:{_id:{city:'$city', state:'$state'}, numberOfzipcodes:{$sum:1}}}, {$sort:{numberOfzipcodes:-1}}, {$group:{_id:'$_id.state', city:{$first:'$_id.city'}, numberOfzipcode:{$first:'$numberOfzipcodes'}}} ]);