arrays - 难以查询具有多个元素的嵌套数组 Mongoose

Question

我已经将文档建模为：

let UserSchema = new Schema({
    name: { type: String, required: true, max: 50, unique: true },
    email: {type: String, required: true, max: 50, unique: true},
    password: {type: String, required: true, max: 20},
    monitoringProject: [
        {
            projectName: { type: String, required: true, max: 30},
            token: { type: String, required: true},
            dispositives: [
                {
                    name: { type: String, required: true, max: 15},
                    value: { type: String, required: true},
                    token: { type: String}
                }
            ]
        }
    ],
    controlProject: [
        {
            projectName: { type: String, required: true, max: 30},
            token: { type: String, required: true},
            dispositives: [
                {
                    name: { type: String, required: true, max: 15},
                    value: { type: String, required: true},
                    token: { type: String}
                }
            ]
        }
    ]
});

例如，在监控项目中，我可以记录太多不同的项目，并且对于每个项目，我可以记录太多的处置。所以，我有嵌套数组。在其他项目中，我只有一个项目和许多处置，我可以列出它们并仅搜索一个处置。但在这种情况下，像这样：

[ { "_id": "5ee18ece1ea5af75bae55f91", "monitoringProject": [ { "dispositives": [ { "_id": "5ee18ef31ea5af75bae55f9c", "name": "Dispositivo 1", "token": "$2b$15$NntmY9ihNZ7R1Vg1WMtmzOSjlfCoKIuHLofF6g5tKXA57WLLc2At2" }, { "_id": "5ee18ef81ea5af75bae55f9e", "name": "Dispositivo 2", "token": "$2b$15$OZB7HbDLvonsv6A71ozW7.yVY2vJOipTCj0AH7XzoafQwwDho.jpm" }, { "_id": "5ee18efc1ea5af75bae55fa0", "name": "Dispositivo 3", "token": "$2b$15$0hpDzNeTO2qJzBVyAHc/UesrnIRi/YpeAlpCLewNhmfl6fe3SEN0i" }, { "_id": "5ee2278d0754817853134bf3", "name": "Dispositivo 4", "token": "$2b$15$ldC5/CyxITPJGUvtXJ.2S.D6a3FBftg6ctQeS4ne/9xpMP5PPiszK" }, { "_id": "5ee2288956fc4c7870044395", "name": "Dispositivo 6", "token": "$2b$15$3rJt.gXEi/gALvrwxVUsWuwF2RUcBmZcoFOQeXwF45ZuKQrB3AGHG" }, { "_id": "5ee6d23d1756b7a8384bcff0", "name": "Dispositivo 7", "token": "$2b$15$C3I9STKJpf9i/7hzhvHEs.IizXpOEmgRTD3hd7ZSdcvATb73Z0zNC" } ], "_id": "5ee18ede1ea5af75bae55f94", "projectName": "Projeto 2", "token": "$2b$15$xaytMbMnGd3BMAGapX8nn.imvkQKetqx0OIlUUyP3gCxlKz/G6DiG" } ] } ]

我只能列出所有的处置。例如，我只想返回“Dispositivo 2”。

我曾经查询：

await User.find({ email: req.body.email, monitoringProject: { $elemMatch: { token: req.body.projecttoken, dispositives: { $elemMatch: { token: req.body.dispositivetoken } } } } }, { "monitoringProject.$.dispositives": 1} )

或者

await User.find({ email: req.body.email, 'monitoringProject.token': req.body.projecttoken, 'monitoringProject.dispositives.token': req.body.dispositivetoken}, { 'monitoringProject.dispositives.$': 1 })

使用 find 或 findOne。我怎样才能只返回我想要的文件？我尝试了很多不同的方法，但没有运气。

标签： arraysmongoosenestedfind