0

这是我的架构：

var A = new Schema({
    active: Boolean
  , containers: [{
         b: { type: ObjectId, ref: 'B' }
    }]
})

var B = new Schema({
    c: { type: ObjectId, ref: 'C' }
  , d: { type: ObjectId, ref: 'D' }
})

var C = new Schema({ })

var D = new Schema({ })

基本上 aA有一个容器数组，这些容器引用了 a B，并且 aB有一个引用 aC和 a D。

现在我有了 a 的 id，C我需要sD使用的一组active A。这甚至可能吗？我应该以某种方式更改架构吗？

编辑：这里的真正架构

 //Mashup has a number of Containers (containerSchema is a sub-doc)
    //so that the same Component could belong to two different Containers
    var containerSchema = new Schema({
          pos: { top: Number, left: Number }
        , size: { width: Number, height: Number }
        , component: { type: ObjectId, ref: 'Component' }
    })

    var mashupSchema = new Schema({
          name: String
        , desc: String
        , size: { width: Number, height: Number }
        , active: Boolean
        , containers: [containerSchema]
    })

    //I am using 'mongoose-schema-extend' to inherit from componentSchema (waiting for the new PR)
    var componentSchema = new Schema({
          name: String
        , desc: String
    }, { collection : 'components', discriminatorKey : '_type' })

    //now the various components
    var imageComponentSchema = componentSchema.extend({
          url: String
    })

    var textComponentSchema = componentSchema.extend({
          text: String
    })


    var htmlComponentSchema = componentSchema.extend({
          html: String
    })

    //this particular component needs a page and a selector
    //(which could live outside it and belong to multiple components)
    var webComponentSchema = componentSchema.extend({
          page: { type: ObjectId, ref: 'Page' }
        , selector: { type: ObjectId, ref: 'Selector' }
    })

    var pageSchema = new Schema({
          name: String
        , desc: String
        , url: String
        , active: { type: Boolean, default: false }
    })

    var selectorSchema = new Schema({
          desc: String
        , url: String
        , cssPath: String
    })

    ///MODELS
    var Mashup = mongoose.model("Mashup", mashupSchema)
    var Component = mongoose.model("Component", componentSchema)
    var ImageComponent = mongoose.model("ImageComponent", imageComponentSchema)
    var TextComponent = mongoose.model("TextComponent", textComponentSchema)
    var HtmlComponent = mongoose.model("HtmlComponent", htmlComponentSchema)
    var WebComponent = mongoose.model("WebComponent", webComponentSchema)
    var Page = mongoose.model("Page", pageSchema)
    var Selector = mongoose.model("Selector", selectorSchema)
4

2 回答 2

1

你觉得太有关系了！现在你，我认为你不能有效地做到这一点，因为你必须：

  • 找到所有有的 Bc: {id}
  • 查找所有具有 a 的 Ab是查询 1 的结果集并且处于活动状态
  • 找出哪些 B 属于您在上一个查询中找到的 ID
  • 找出哪些 D 属于你找到的 B

我认为你绝对应该在这里非规范化你的模式。例如，您可以将以上所有内容放在一个文档中：

{
    Active: true,
    containers: [ // B's
        { c: [
            { _id: X, field1: foo },
            { _id: X, field1: foo },
        ] },
        { d: [
            { _id: X, field1: foo },
            { _id: X, field1: foo },
        ] }
    ]
}

然后你可以用一个查询来做到这一点：

db.collection.find(
    { "container.c._id" : your_id, Active: true }, // query
    { "container.d" : 1 } // projection
);
于 2013-08-15T14:35:08.933 回答
0

最终完全改变了数据库模式。关闭。

于 2013-08-21T12:35:45.697 回答