好像你是依靠每个人的首选阵列中最喜欢的食物。如果是这样,那么您可以利用一个聚合框架运算符。
这里是管道可以使用:
db.people.aggregate(
[
{
"$unwind" : "$favourite_foods"
},
{
"$group" : {
"_id" : {
"name" : "$name",
"height" : "$height"
},
"faveFood" : {
"$first" : "$favourite_foods"
}
}
},
{
"$group" : {
"_id" : "$faveFood.name",
"height" : {
"$max" : "$_id.height"
}
}
}
])
这个样本数据集:
> db.people.find().pretty()
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 6
}
{
"_id" : ObjectId("5088950bd4197aa2b9490742"),
"name" : "Lucy",
"favourite_foods" : [
{
"name" : "Pasta",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 5.5
}
{
"_id" : ObjectId("5088951dd4197aa2b9490743"),
"name" : "Landy",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Pizza",
"type" : "Veggie"
}
],
"height" : 5
}
{
"_id" : ObjectId("50889541d4197aa2b9490744"),
"name" : "Augie",
"favourite_foods" : [
{
"name" : "Sushi",
"type" : "Four Cheeses"
},
{
"name" : "Pizza",
"type" : "Veggie"
}
],
"height" : 6.2
}
你得到这些结果:
{
"result" : [
{
"_id" : "Pasta",
"height" : 5.5
},
{
"_id" : "Pizza",
"height" : 6
},
{
"_id" : "Sushi",
"height" : 6.2
}
],
"ok" : 1
}
是的,我发现了。我有点希望有人提出了一个解决方法... – Russell
你不能提取位置的任意元素,你碰巧有办法通过$ unwind获取第一个或最后一个元素,并用$ first或$ last操作符重新获得$ group。 –
**好消息**:该问题最终通过[arrayElementAt运算符](https://docs.mongodb.com/manual/reference/operator/aggregation/arrayElemAt/) – joeytwiddle