Question

实际上我并不是想要匹配两个系列。我正在寻找与二次收藏相匹配的领域。我会清楚地解释我的问题：这是我的主要收藏品“bags”。其中包含

 {
    "_id" : ObjectId("568f43e08a9f71b70b22a694"),
            "bagNo" : "HBBN/00001/16",
            "category" : "Voluntary",
            "regNo" : "DNR/00001/16",
            "status": "Accepted"
    }, 
{ "_id" : ObjectId("568f4645fa0758af0e3fef26"), "bagNo" :"HBBN/00002/16", "category" : "Voluntary", "regNo" : "DNR/00002/16", "status": "Rejected" },

{ "_id" : ObjectId("568f4645fa4546gygfef26"), "bagNo" : "HBBN/00003/16", "category" : "Voluntary", "regNo" : "DNR/00003/16", "status": "Accepted" }

现在我的加入集合“捐赠者”包含

 {"donorType" : "H",
     "regNo" : "DNR/00001/16",
     "surName" : "pandey",
     "firstName" : "rakesh"}, 
{"donorType" : "C", "regNo" : "DNR/00002/16", "surName" : "pandey", "firstName" : "rakesh" },

{"donorType" : "C", "regNo" : "DNR/00003/16", "surName" : "pandey", "firstName" : "rakesh" }

从这两个系列我想从袋子收集中获得已接受的袋子数据，在捐赠者收藏中将'donortype'作为'H'。如果有可能，请告诉我。解释一些例子

Answer 1

要合并两个集合，您可以尝试在聚合中使用$lookup。

donor.aggregation([
      // filter the `donortype` of `H`
      {$match: {donortype: 'H'}},
      // join with `bags` collection for the foreign key `regNo`, and put bags into `h_bags` result. 
      {$lookup: {
          from: "bags",
          localField: "regNo",
          foreignField: "regNo",
          as: "h_bags"
        }},
      // unwind the `h_bags` array
      {$unwind: '$h_bags'},
      // filter the status of bags is `Accepted`
      {$match: {'h_bags.status': 'Accepted'}}
   ]);

Mongodb使用多个键语句连接聚合

1 个答案: