Mongoose - 按标准查找子文档 [英] Mongoose - finding subdocuments by criteria

问题描述

我刚刚被这个问题困住了.我有两个猫鼬模式:

I've just got stuck with this problem. I've got two Mongoose schemas:

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

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

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

Question is, how to fetch all subdocuments (in this case, childrenSchema objects) from every parent document? Let's suppose I have some data:

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 }
    ]}
];

我想在单个查询中检索所有 18 岁以上的儿童.这可能吗?每个答案都将不胜感激，谢谢！

I would like to retrieve - in a single query - all children older than 18. Is it possible? Every answer will be appreciated, thanks!

推荐答案

您可以在最新的 MongoDB 版本中使用 $elemMatch 作为查询投影运算符.来自 mongo shell:

You can use $elemMatch as a query-projection operator in the most recent MongoDB versions. From the mongo shell:

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

这会从 children 数组中过滤掉年幼儿童的文档:

This filters younger children's documents out of the children array:

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

如您所见，子文档仍分组在其父文档中.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:

As you can see, children are still grouped inside their parent documents. MongoDB queries return documents from collections. You can use the aggregation framework's $unwind method to split them into separate documents:

> 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
}

为了性能，我重复了 $match 子句:第一次通过它消除了没有孩子至少 18 岁的父母，所以 $unwind 只考虑有用的文件.第二个 $match 删除不匹配的 $unwind 输出，$project 将子文档中的子文档提升到顶级.

I repeat the $match clause for performance: the first time through it eliminates parents with no children at least 18 years old, so the $unwind only considers useful documents. The second $match removes $unwind output that doesn't match, and the $project hoists children's info from subdocuments to the top level.

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