Java 按投影字段分组时引用无效

Java 按投影字段分组时引用无效,java,mongodb,spring-data,spring-data-mongodb,Java,Mongodb,Spring Data,Spring Data Mongodb,我有两个文档:比如说Foo和Qux Foo如下所示: { "_id": ObjectId("5c52bb1af9b7bb512458a6d1"), "name": "Foo 1", "description": "This is a Foo", "bars": [ { "name": "Bar 1", "description": "This is a Bar", "qux"

我有两个文档:比如说
Foo
Qux

Foo
如下所示:

{
    "_id": ObjectId("5c52bb1af9b7bb512458a6d1"),
    "name": "Foo 1",
    "description": "This is a Foo",
    "bars": [ 
        {
            "name": "Bar 1",
            "description": "This is a Bar",
            "qux": ObjectId("5c3f3d59d45cca2d1860bb4e")
        },         
        {
            "name": "Bar 2",
            "description": "This is a Bar",
            "qux": ObjectId("5c3f3d59d45cca2d1860bb4e")
        }
    ]
}
Qux
看起来像:

{
    "_id": ObjectId("5c3f3d59d45cca2d1860bb4e"),
    "name": "Qux 1",
    "description": "This is a Qux"
}
我的目标是将相应的
Qux
嵌入
Foo.bar
的每个元素,如下所示:

[{
    "_id": ObjectId("5c52bb1af9b7bb512458a6d1"),
    "name": "Foo 1",
    "description": "This is a Foo",
    "bars": [ 
        {
            "name": "Bar 1",
            "description": "This is a Bar",
            "qux": {
                "_id": ObjectId("5c3f3d59d45cca2d1860bb4e"),
                "name": "Qux 1",
                "description": "This is a Qux"
             }
        },         
        {
            "name": "Bar 2",
            "description": "This is a Bar document",
            "qux": {
                "_id": ObjectId("5c3f3d59d45cca2d1860bb4e"),
                "name": "Qux 1",
                "description": "This is a Qux"
             }
        }
    ]
}]
我尝试了以下聚合:

Aggregation agg = Aggregation.newAggregation(
    Aggregation.match(Criteria.where("_id").is(id)),
    Aggregation.unwind("bars", true),
    Aggregation.lookup("qux", "bars.qux", "_id", "bars.qux"),
    Aggregation.project("_id", "name", "description")
        .and("bars.qux").arrayElementAt(0).as("bars.qux")
        .and("bars.name").as("bars.name")
        .and("bars.description").as("bars.description"),
    Aggregation.group("_id")
        .push("bars").as("bars")
        .first("name").as("name")
        .first("description").as("description")
);
但是由于这一行
.push(“bar”).as(“bar”)
,它抛出了一个
IllegalArgumentException

如果我在不执行组操作的情况下执行聚合,它会工作,但我会为每个bar元素获取一个Foo,并且每个Foo包含一个不同的bar元素,这是我所期望的,因为我将它们展开:

[{
    "_id": ObjectId("5c52bb1af9b7bb512458a6d1"),
    "name": "Foo 1",
    "description": "This is a Foo",
    "bars": {
        "name": "Bar 1",
        "description": "This is a Bar",
        "qux": {
            "_id": ObjectId("5c3f3d59d45cca2d1860bb4e"),
            "name": "Qux 1",
            "description": "This is a Qux"
         }
    }
},
{
    "_id": ObjectId("5c52bb1af9b7bb512458a6d1"),
    "name": "Foo 1",
    "description": "This is a Foo",
    "bars": {
        "name": "Bar 2",
        "description": "This is a Bar",
        "qux": {
            "_id": ObjectId("5c3f3d59d45cca2d1860bb4e"),
             "name": "Qux 1",
             "description": "This is a Qux"
        }
    }
}]

有什么方法可以实现我的目标吗?

您可以在不使用
$unwind
的情况下获得所需的输出。一旦我们
$lookup
我们就可以
$map
qux rom通过
$indexOfArray
$arrayElemAt
加入数组,并使用
$mergeObjects
合并对象

db.foo.aggregate([
    {$lookup: {from : "qux", localField : "bars.qux", foreignField : "_id", as : "join"}},
    {$addFields: {bars: {$map : {input : "$bars", as : "b", in : {$mergeObjects :[ "$$b", {qux: {$arrayElemAt : ["$join", {$indexOfArray: ["$join._id", "$$b.qux"]}]}}]}}}}},
    {$project: {join:0}}
]).pretty()
输出

{
        "_id" : ObjectId("5c52bb1af9b7bb512458a6d1"),
        "name" : "Foo 1",
        "description" : "This is a Foo",
        "bars" : [
                {
                        "name" : "Bar 1",
                        "description" : "This is a Bar",
                        "qux" : {
                                "_id" : ObjectId("5c3f3d59d45cca2d1860bb4e"),
                                "name" : "Qux 1",
                                "description" : "This is a Qux"
                        }
                },
                {
                        "name" : "Bar 2",
                        "description" : "This is a Bar",
                        "qux" : {
                                "_id" : ObjectId("5c3f3d59d45cca2d1860bb4e"),
                                "name" : "Qux 1",
                                "description" : "This is a Qux"
                        }
                }
        ]
}

这是可行的,但我需要在java中用spring数据mongodb编写它。
{
        "_id" : ObjectId("5c52bb1af9b7bb512458a6d1"),
        "name" : "Foo 1",
        "description" : "This is a Foo",
        "bars" : [
                {
                        "name" : "Bar 1",
                        "description" : "This is a Bar",
                        "qux" : {
                                "_id" : ObjectId("5c3f3d59d45cca2d1860bb4e"),
                                "name" : "Qux 1",
                                "description" : "This is a Qux"
                        }
                },
                {
                        "name" : "Bar 2",
                        "description" : "This is a Bar",
                        "qux" : {
                                "_id" : ObjectId("5c3f3d59d45cca2d1860bb4e"),
                                "name" : "Qux 1",
                                "description" : "This is a Qux"
                        }
                }
        ]
}