MongoDB：如何获取子文档字段值的不同列表？
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MongoDB: How to get distinct list of sub-document field values?
提问by vladimir
Let's say I have the following documents in collection:
假设我收集了以下文件：
{
"family": "Smith",
"children": [
{
"child_name": "John"
},
{
"child_name": "Anna"
},
]
}
{
"family": "Williams",
"children": [
{
"child_name": "Anna"
},
{
"child_name": "Kevin"
},
]
}
Now I want to get somehow the following list of unique child names cross all families:
现在我想以某种方式获得以下所有家庭的唯一儿童姓名列表：
[ "John", "Anna", "Kevin" ]
Structure of result might be different. How to achieve that in MongoDB? Should be something simple but I can't figure out. I tried aggregate() function on collection but then I don't know how to apply distinct() function.
结果的结构可能不同。如何在 MongoDB 中实现这一目标？应该很简单，但我无法弄清楚。我在集合上尝试了 aggregate() 函数，但后来我不知道如何应用 distinct() 函数。
回答by Asya Kamsky
You can just do:
你可以这样做：
db.collection.distinct("children.child_name");
In your case it returns:
在您的情况下，它返回：
[ "John", "Anna", "Kevin" ]
回答by Vladimir Muzhilov
with the help aggregation framework:
使用帮助聚合框架：
db.collection.aggregate([{$unwind:'$children'}, {$group:{_id:'$children.child_name'}}])
or more interest ;) with frequency of name:
或更多的兴趣 ;) 与名字的频率：
db.collection.aggregate([{$unwind:'$children'}, {$group:{_id:'$children.child_name', freq:{$sum:1}}}])
