Question

我的藏品中有以下文件

{
_id: ObjectId("5166fefbc482c31052000002"),
contact: [
    {
        home: 7735734105
    },
    {
        office: 9583866301
    }
],
user_name: "moti",
reportsTo: "bikram",
technology: [
    {
        name: ".net",
        rating: 5
    },
    {
        name: "JavaScript",
        rating: 2
    }
],
project: [
    "Agile School",
    "Draftmate"
],
type: "developer",
email: "motiranjan.pradhan@ajatus.co.in"

} 和

{
_id: ObjectId("5166fe90c482c31052000001"),
contact: [
    {
        home: 7735734103
    },
    {
        office: 9583866901
    }
],
user_name: "ganesh",
reportsTo: "bikram",
technology: [
    {
        name: "JavaScript",
        rating: 3
    },
    {
        name: ".net",
        rating: 4
    }
],
project: [
    "SLBC",
    "Draftmate"
],
type: "developer",
email: "ganesh.patra@ajatus.co.in"

}

现在我需要找到只知道JavaScript的人的评级。目前我跑

db.users.find( { technology: { $elemMatch: { name: 'JavaScript' } } },{user_name:1,'technology.name':1,_id:0} ).pretty()

我正在获取所有技术（.net和JavaScript）的名称及其相应的评级。我只需要用户名和他们各自的JavaScript评级。我需要使用任何聚合技术吗？

Answer 1

positional operator '$'可用于将查询结果限制为第一个匹配元素。要在上面的查询中使用，您可以将其更改为：

db.users.find( { technology: { $elemMatch: { name: 'JavaScript' } } },{user_name:1,'technology.$.name':1,_id:0} )

使用mongoDB find获取所需的输出

1 个答案: