Question

我将使用here

中的示例

{
 _id: 1,
 zipcode: 63109,
 students: [
              { name: "john", school: 102, age: 10 },
              { name: "jess", school: 102, age: 11 },
              { name: "jeff", school: 108, age: 15 }
           ]
}
{
 _id: 2,
 zipcode: 63110,
 students: [
              { name: "ajax", school: 100, age: 7 },
              { name: "achilles", school: 100, age: 8 },
           ]
}

{
 _id: 3,
 zipcode: 63109,
 students: [
              { name: "ajax", school: 100, age: 7 },
              { name: "achilles", school: 100, age: 8 },
           ]
}

{
 _id: 4,
 zipcode: 63109,
 students: [
              { name: "barney", school: 102, age: 7 },
           ]
}

如果我跑

db.schools.find( { zipcode: 63109 },
             { students: { $elemMatch: { school: 102 } } } )

它将给出每个数组的第一个结果。这个命名：

{ "_id" : 1, "students" : [ { "name" : "john", "school" : 102, "age" : 10 } ] }
{ "_id" : 3 }
{ "_id" : 4, "students" : [ { "name" : "barney", "school" : 102, "age" : 7 } ] }

如何让它返回符合条件的数组的所有对象（而不仅仅是第一个）？这意味着：

{
 _id: 1,
 students: [
              { name: "john", school: 102, age: 10 },
              { name: "jess", school: 102, age: 11 }
           ]
}    
{ _id: 3 }
{_id: 4, students: [ { name: "barney", school: 102, age: 7 }]}

Answer 1

为了返回多个子文档，您将需要使用聚合框架。这将返回您正在寻找的所有子文档：

db.zip.aggregate(
  {$match: {zipcode: 63109}},
  {$unwind: "$students"},
  {$match: {"students.school": 102}}
)

你可以做各种事情来获得不同的输出，但这将返回：

{
    "result" : [
        {
            "_id" : 1,
            "zipcode" : 63109,
            "students" : {
                "name" : "john",
                "school" : 102,
                "age" : 10
            }
        },
        {
            "_id" : 1,
            "zipcode" : 63109,
            "students" : {
                "name" : "jess",
                "school" : 102,
                "age" : 11
            }
        },
        {
            "_id" : 4,
            "zipcode" : 63109,
            "students" : {
                "name" : "barney",
                "school" : 102,
                "age" : 7
            }
        }
    ],
    "ok" : 1
}

Answer 2

以前和错误答案：

~~这应该适用于今天。见https://docs.mongodb.com/v3.2/reference/operator/projection/positional/#array-field-limitations~~

在查询中使用$elemMatch查询并在投影中公开子文档时，您应该得到正确的结果，如下所示：

db.schools.find( { zipcode: 63109, students: { $elemMatch: { school: 102 } } }, { 'students.$': 1 } )

新答案

使用find()，现在无法将子文档列表限制为与查询匹配的子文档。请改为aggregate()或采取以下其中一种方式：

您可以通过在投影中添加array-property来获取匹配文档的所有子文档：

db.schools.find( { zipcode: 63109, students: { $elemMatch: { school: 102 } } }, { 'students': 1 })
> { "_id" : 1, "students" : [ { "name" : "john", "school" : 102, "age" : 10 }, { "name" : "jess", "school" : 102, "age" : 11 }, { "name" : "jeff", "school" : 108, "age" : 15 } ] }
> { "_id" : 4, "students" : [ { "name" : "barney", "school" : 102, "age" : 7 } ] }

或者您可以在子文档中获得与$elemMatch查询匹配的第一个项目：

db.schools.find( { zipcode: 63109, students: { $elemMatch: { school: 102 } } }, { 'students.$': 1 })
> { "_id" : 1, "students" : [ { "name" : "john", "school" : 102, "age" : 10 } ] }
> { "_id" : 4, "students" : [ { "name" : "barney", "school" : 102, "age" : 7 } ] }

Answer 3

这在类似过滤的情况下对我有用。现在，我知道这个问题是很多年前问过的，但是对任何想要像我这样的答案的人来说都是这样。这对我有用。多亏了原始答案！

对于此特定问题： outerparam是zipcode，而innerarray.property是students.school。

let cursor = db
    .collection("somecollection")
    .aggregate(
      { $match: { outerparam: outermatch } },
      { $unwind: "$innerarray" },
      { $match: { "innerarray.property": propertymatch } },
      { $project: { "innerarray.$": 1 } });

使$ elemMatch（投影）返回符合条件的所有对象

3 个答案: