MongoDB 不同聚合
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MongoDB distinct aggregation
提问by Lemonio
I'm working on a query to find cities with most zips for each state:
我正在查询以查找每个州的邮编最多的城市：
db.zips.distinct("state", db.zips.aggregate([ {$group:{_id:{state:"$state", city:"$city"},numberOfzipcodes:{$sum:1}}}, {$sort:{numberOfzipcodes:-1}}]))
The aggregate part of the query seems to work fine, but when I add the distinct I get an empty result.
查询的聚合部分似乎工作正常，但是当我添加 distinct 时，我得到一个空结果。
Is this because I have state in the id? Can I do something like distinct("_id.state?
这是因为我在 id 中有状态吗？我可以做类似的事情distinct("_id.state吗？
回答by dam1
回答by Sammaye
Distinct and the aggregation framework are not inter-operable.
Distinct 和聚合框架不可互操作。
Instead you just want:
相反，您只想要：
db.zips.aggregate([
{$group:{_id:{city:'$city', state:'$state'}, numberOfzipcodes:{$sum:1}}},
{$sort:{numberOfzipcodes:-1}},
{$group:{_id:'$_id.state', city:{$first:'$_id.city'},
numberOfzipcode:{$first:'$numberOfzipcodes'}}}
]);
回答by Surendranath Reddy K
SQL Query: (group by & count of distinct)
SQL 查询：（分组依据和不同计数）
select city,count(distinct(emailId)) from TransactionDetails group by city;
Equivalent mongo query would look like this:
等效的 mongo 查询如下所示：
db.TransactionDetails.aggregate([
{$group:{_id:{"CITY" : "$cityName"},uniqueCount: {$addToSet: "$emailId"}}},
{$project:{"CITY":1,uniqueCustomerCount:{$size:"$uniqueCount"}} }
]);
回答by Jeroen
You can call $setUnionon a single array, which also filters dupes:
您可以调用$setUnion单个数组，该数组还可以过滤重复项：
{ $project: {Package: 1, deps: {'$setUnion': '$deps.Package'}}}
