Question

我有几个solutions，其中可以容纳多个projects。我通过embedding解决方案文档中的项目来建模这种关系，例如

[{
    _id: "1",
    solutionTitle:  "Some Test Solution",        
    projects: [
      {
         _id: "12",
         type: "Java",
         title: "Test Project"
      },
      {
         _id: "13",
         type: "Misc",
         title: "Test Project"
      }
    ]
 },
 {
    _id: "2",
    solutionTitle:  "A Different Solution",  
    projects: [
      {
         _id: "21",
         type: "Java",
         title: "Another Java Project"
      }
    ]
 }]

现在我想select all projects特殊类型，例如Java。我使用聚合尝试了以下query：

db.Solutions.aggregate ( 
    { "$unwind": "$projects" }, 
    { "$match": {"projects.type": "Java" } }, 
    { "$project": {"projects" : 1, "_id": 0, "solutionTitle": 0 } } 
)

这很好但结果看起来不像我的预期。我得到了

{
    projects: {
        _id: "12",
        type: "Java",
        title: "Test Project"
    },
    projects: {
        _id: "21",
        type: "Java",
        title: "Another Java Project"
    }
}

如何将结果作为项目列表，例如

[
    { _id: "12", type: "Java", title: "Test Project" }
    ...
]

我检查了this SO question和this one，但它们并没有真正涵盖我需要的内容。非常感谢任何帮助。

Answer 1

如果您希望它们位于数组中，则需要将它们添加到数组中：

db.Solutions.aggregate ([
    { "$unwind": "$projects" }, 
    { "$match": {"projects.type": "Java" } }, 
    { "$group": {"_id": null, "projects": { "$push": "$projects" } } }
])

根据您在下面的评论，您真正想要的是返回项目子文档，就好像它们是您正在搜索的文档一样。在这种情况下，您应明确$project您的项目值：

db.Solutions.aggregate ([
    { "$unwind": "$projects" }, 
    { "$match": {"projects.type": "Java" } }, 
    { "$project": {"_id": "$projects._id", "type": "$projects.type", "title": "$projects.title" } }
])

MongoDB：嵌入式文档的查询和返回列表

1 个答案: