我正在try 聚合以下文档,以将参与者作为嵌套数组中的对象.
{
"name": "EXAMPLE",
"schedules": [
{
"schedule_id": "id1",
"participants": [
"participant_id1",
"participant_id2"
],
},
{
"schedule_id": "id2",
"participants": [
"participant_id1",
"participant_id2"
],
},
{
"schedule_id": "id3",
"participants": [
"participant_id1"
],
},
],
}
因此,我写了以下管道:
[
{
$unwind: {
path: "$schedules",
includeArrayIndex: "index",
preserveNullAndEmptyArrays: true,
}
},
{
$unwind: {
path: "$schedules.participants",
includeArrayIndex: "index",
preserveNullAndEmptyArrays: true,
}
},
{
$lookup: {
from: "customers",
localField: "schedules.participants",
foreignField: "_id",
as: "participants",
}
},
{
$project: {
"participants.address": 0,
"participants.birthday": 0,
}
},
{
$unwind: {
path: "$participants",
preserveNullAndEmptyArrays: true,
}
},
{
$group:
{
_id: "$_id",
name: {
$first: "$name",
},
schedules: {
$first: "$schedules",
},
}
},
]
- 此管道中的第一步是展开时间表数组,以获得文档中的每个单独时间表.
- 第二步是展开参与者,因为我需要参与者ID来执行第三步中的查找过程.
- 第三步是在Customers集合中查找参与者,返回的将是一个Customer对象.
- 在第四步中,我将使用project从给定参与者中删除不必要的字段.
- 在第五步中,我再次使用展开来获取单个参与者(我知道也可以使用$First运算符)
- 在第六步,我将对其进行分组
我正在try 将步骤3中的每个参与者添加到参与者数组中的相应日程对象,即document should be like this:
{
"name": "EXAMPLE",
"schedules": [
{
"schedule_id": "id1",
"participants": [
{
id: "id1",
"name": "name1"
},
{
id: "id2",
"name": "name2"
},
],
},
{
"schedule_id": "id2",
"participants": [
{
id: "id1",
"name": "name1"
},
{
id: "id2",
"name": "name2"
},
],
},
{
"schedule_id": "id3",
"participants": [
{
id: "id1",
"name": "name1"
},
],
},
],
}