如何在MongoDB中比较聚合过滤器中的两个字段？
为此，将aggregate()与$filter一起使用。让我们创建一个包含文档的集合-
> db.demo137.insertOne(
... {
... Name1:"Chris",
... Name2:"David",
... Detail1:[
... {_id:"John", Name3:"Chris"},
... {_id:"Chris", Name3:"Chris"},
... ],
... Detail2:[{_id:"David", Name3:"Chris"},
... {_id:"Carol", Name3:"Chris"},
... {_id:"John", Name3:"Chris"}]
... }
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5e31b4cefdf09dd6d08539a0")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo137.find();
这将产生以下输出-
{
"_id" : ObjectId("5e31b4cefdf09dd6d08539a0"), "Name1" : "Chris", "Name2" : "David", "Detail1" : [
{ "_id" : "John", "Name3" : "Chris" },
{ "_id" : "Chris", "Name3" : "Chris" } ], "Detail2" : [ { "_id" : "David", "Name3" : "Chris" },
{ "_id" : "Carol", "Name3" : "Chris" }, { "_id" : "John", "Name3" : "Chris" }
]
}以下是查询以比较聚合过滤器中的两个字段-
> db.demo137.aggregate([
... {
... $project: {
... _id: 0,
... Name1: 1,
... Name2: 2,
... Detail1: {
... $filter: {
... input: "$Detail1",
... as: "out",
... cond: {
... $eq: ["$$out._id", "$Name1" ]
... }
... }
... },
... Detail2: {
... $filter: {
... input: "$Detail2",
... as: "out",
... cond: {
... $eq: ["$$out._id", "$Name2" ]
... }
... }
... }
... }
... }
... ])这将产生以下输出-
{
"Name1" : "Chris", "Name2" : "David", "Detail1" : [
{ "_id" : "Chris", "Name3" : "Chris" } ], "Detail2" : [ { "_id" : "David", "Name3" : "Chris" }
]
}