MongoDB: Query For and Return List of Embedded Documents

Question

I have several solutions, which in term may hold several projects. I modeled this relationship by embedding the projects in the solution document, e.g.

[{
    _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"
      }
    ]
 }]

Now I want to select all projects of a special type, e.g. Java. I tried the following query with aggregation:

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

This works fine but the result doesn't look like what I expected. I get

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

How can I get the result to be a list of projects, e.g.

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

I checked this SO question and this one, but they do not really cover what I need. Any help is greatly appreciated.

NoOutlet · Accepted Answer · 2015-01-27 12:34:50Z

1

If you want them to be in an array, you'll need to add them to an array:

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

In light of your comments below, it seems what you really want is to return the project subdocuments as if they were the documents you were searching. In which case you should $project your project values explicitly:

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

edited Jan 27, 2015 at 12:34

answered Jan 27, 2015 at 12:13

NoOutlet

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4 Comments

WeSt Over a year ago

Can I get it without the object wrapper as well? Just a list of projects directly? Now I get an object with a property projects, which is a list.

NoOutlet Over a year ago

I have updated my answer to satisfy that requirement.

WeSt Over a year ago

Thx! Is there a way to take all fields automatically? This is rather tiresome if you have a lot of columns.

NoOutlet Over a year ago

I don't believe there is, but I'd be happy to be proved wrong on that.

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