Mongoose - 按标准查找子文档

Question

我刚刚遇到这个问题.我有两个Mongoose模式:

var childrenSchema = mongoose.Schema({
    name: {
        type: String
    },
    age: {
        type: Number,
        min: 0
    }
});

var parentSchema = mongoose.Schema({
    name : {
        type: String
    },
    children: [childrenSchema]
});

问题是,如何childrenSchema从每个父文档中获取所有子文档(在本例中为对象)？我们假设我有一些数据:

var parents = [
    { name: "John Smith",
    children: [
        { name: "Peter", age: 2 }, { name: "Margaret", age: 20 }
    ]},
    { name: "Another Smith",
    children: [
        { name: "Martha", age: 10 }, { name: "John", age: 22 }
    ]}
];

我想在一个查询中检索所有18岁以上的孩子.这可能吗？每个答案都将不胜感激,谢谢!

Answer 1

您可以$elemMatch在最新的MongoDB版本中用作查询 - 投影运算符.来自mongo shell:

db.parents.find(
    {'children.age': {$gte: 18}},
    {children:{$elemMatch:{age: {$gte: 18}}}})

这会将年幼儿童的文档从children数组中过滤出来:

{ "_id" : ..., "children" : [ { "name" : "Margaret", "age" : 20 } ] }
{ "_id" : ..., "children" : [ { "name" : "John", "age" : 22 } ] }

如您所见,儿童仍然在其父文档中进行分组.MongoDB查询从集合中返回文档.您可以使用聚合框架的$unwind方法将它们拆分为单独的文档:

> db.parents.aggregate({
    $match: {'children.age': {$gte: 18}}
}, {
    $unwind: '$children'
}, {
    $match: {'children.age': {$gte: 18}}
}, {
    $project: {
        name: '$children.name',
        age:'$children.age'
    }
})
{
    "result" : [
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb5"),
            "name" : "Margaret",
            "age" : 20
        },
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb6"),
            "name" : "John",
            "age" : 22
        }
    ],
    "ok" : 1
}

我重复了$match表现的条款:第一次通过它取消了没有孩子至少18岁的父母,所以$unwind只考虑有用的文件.第二个$match删除$unwind不匹配的输出,并将$project子文档中的子项信息提升到顶层.

Answer 2

在Mongoose中,您还可以使用这样的优雅.populate()功能:

parents
.find({})
.populate({
  path: 'children',
  match: { age: { $gte: 18 }},
  select: 'name age -_id'
})
.exec()