如果MongoDB中至少有一个元素匹配，如何聚合两个列表？
> db.demo456.insertOne(
... { _id: 101, StudentName: ["Chris", "David"] }
... );
{ "acknowledged" : true, "insertedId" : 101 }
>
> db.demo456.insertOne(
... { _id: 102, StudentName: ["Mike", "Sam"] }
... );
{ "acknowledged" : true, "insertedId" : 102 }
> db.demo456.insertOne(
... { _id: 103, StudentName: ["John", "Jace"] }
... );
{ "acknowledged" : true, "insertedId" : 103 }
> db.demo456.insertOne(
... { _id: 104, StudentName: ["Robert", "John"] }
... );
{ "acknowledged" : true, "insertedId" : 104 }在find()方法的帮助下显示集合中的所有文档-
> db.demo456.find();
这将产生以下输出-
{ "_id" : 101, "StudentName" : [ "Chris", "David" ] }
{ "_id" : 102, "StudentName" : [ "Mike", "Sam" ] }
{ "_id" : 103, "StudentName" : [ "John", "Jace" ] }
{ "_id" : 104, "StudentName" : [ "Robert", "John" ] }以下是如果至少一个元素匹配则汇总两个列表的查询-
> db.demo456.aggregate([
... {$unwind:"$StudentName"},
... {$group:{_id:"$StudentName", combine:{$addToSet:"$_id"}, size:{$sum:1}}},
... {$match:{size: {$gt: 1}}},
... {$project:{_id: 1, combine:1, size: 1, combine1: "$combine"}},
... {$unwind:"$combine"},
... {$unwind:"$combine1"},
... {$group:{_id:"$combine", l:{$first:"$_id"}, size:{$sum: 1}, set: {$addToSet:"$combine1"}}},
... {$sort:{size:1}},
... {$group:{_id: "$l", combineIds:{$last:"$set"}, size:{$sum:1}}},
... {$match: {size:{$gt:1}}}
... ])这将产生以下输出-
{ "_id" : "John", "combineIds" : [ 103, 104 ], "size" : 2 }热门推荐
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