mongodb 如何在MongoDB中汇总总和以获得总数？
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How to aggregate sum in MongoDB to get a total count?
提问by Sahat Yalkabov
For some collection with a field { wins: Number }, how could I use MongoDB Aggregation Frameworkto get the total number of wins across all documents in a collection?
对于某些带有字段的集合，{ wins: Number }我如何使用MongoDB 聚合框架来获取集合中所有文档的总获胜次数？
Example:
例子：
If I have 3 documents with wins: 5, wins: 8, wins: 12respectively, how could I use MongoDB Aggregation Framework to return the total number, i.e. total: 25.
如果我有3个文件wins: 5，wins: 8，wins: 12分别，我怎么可能使用MongoDB的聚合框架返回的总数，即total: 25。
回答by WiredPrairie
Sum
和
To get the sum of a grouped field when using the Aggregation Framework of MongoDB, you'll need to use $groupand $sum:
要在使用 MongoDB 的聚合框架时获取分组字段的总和，您需要使用$group和$sum：
db.characters.aggregate([ {
$group: {
_id: null,
total: {
$sum: "$wins"
}
}
} ] )
In this case, if you want to get the sum of all of the wins, you need to refer to the field name using the $syntax as $winswhich just fetches the values of the winsfield from the grouped documents and sums them together.
在这种情况下，如果您想获得所有 的总和wins，您需要使用$语法引用字段名称，因为$wins它只是wins从分组文档中获取字段的值并将它们相加。
Count
数数
You can sumother values as well by passing in a specific value (as you'd done in your comment). If you had
您也可以sum通过传入特定值来获取其他值（就像您在评论中所做的那样）。如果你有
{ "$sum" : 1 },
{ "$sum" : 1 },
that would actually be a count of all of the wins, rather than a total.
这实际上是所有 的计数wins，而不是总数。
