mongodb-基于条件字段合并两个对象
假设我们有以下由聚合管道生成的文档:mongodb-基于条件字段合并两个对象,mongodb,mongodb-query,aggregation-framework,Mongodb,Mongodb Query,Aggregation Framework,假设我们有以下由聚合管道生成的文档: [ { "_id": ObjectId("5a934e000102030405000000"), "description": "description for item 1", "item_code": "00001" }, { "_id": ObjectId("5
[
{
"_id": ObjectId("5a934e000102030405000000"),
"description": "description for item 1",
"item_code": "00001"
},
{
"_id": ObjectId("5a934e000102030405000001"),
"description": "description for item 2",
"item_code": "00002"
},
{
"_id": ObjectId("5a934e000102030405000002"),
"description": "description for item 3",
"item_code": "00003"
},
{
"_id": ObjectId("5a934e000102030405000003"),
"extrafield": "extra field for item 2",
"item_code": "00002"
}
]
如何将具有相同项目\u code
的文档合并为一个文档,并保留所有属性?
预期结果:
[
{
"description": "description for item 1",
"item_code": "00001"
},
{
"description": "description for item 2",
"extrafield": "extra field for item 2",
"item_code": "00002"
},
{
"description": "description for item 3",
"item_code": "00003"
}
]
我尝试了不同的$group
模式,但没有成功:(
你可以试试
按$group
,使用项_code
与$mergeObjects
合并对象$$ROOT
将根对象替换为根对象$replaceWith
谢谢@turivishal。我正朝着这个方向前进,但使用$push:“$$ROOT”(而不是$mergeObjects:“$$ROOT”)却没有成功。老实说,我不明白为什么你的解决方案有效,而我的解决方案无效:)按id分组意味着,组拥有所有分组对象,
$mergeObjects
将在单个对象中合并对象。
db.collection.aggregate([
{
$group: {
_id: "$item_code",
root: { $mergeObjects: "$$ROOT" }
}
},
{ $replaceWith: "$root" }
])