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[英]How to sort Arrays of Object using aggregation in mongoDB
[英]Compare arrays and filter, using MongoDB aggregation
对于我的数据库,我编写了以下管道:
let orders = await Order.aggregate(
{
$unwind: "$candidate",
},
{
$lookup: {
from: "groups",
localField: "candidate.groupId",
foreignField: "_id",
as: "groupData",
},
},
{
$lookup: {
from: "users",
let: {
id: "$candidate.groupId",
},
pipeline: [
{ $match: { groupId: { $ne: null } } },
{
$match: {
$expr: { $in: ["$$id", "$groupId"] },
},
},
{ $project: { name: 1, email: 1, _id: 1 } },
],
as: "members",
},
},
{ $match: { "members._id": new ObjectId(req.userId) } },
{
$lookup: {
from: "users",
let: { ids: "$candidate.autonomousId" },
pipeline: [
{
$match: {
$expr: { $in: ["$_id", "$$ids"] },
},
},
{ $project: { name: 1, email: 1, _id: 1 } },
],
as: "candidate",
},
},
{
$project: {
groupData: 1,
members: 1,
candidate: 1,
stillAvailable: 1,
_id: 0,
},
}
).toArray();
output 是预期的……
{ candidate:
[ { _id: 601817dc2eeecd17db3a68f6,
name: 'Maria' },
{ _id: 601817ef2eeecd17db3a68f7,
name: 'Jose' } ],
groupData:
[ { _id: 606632403fffb851b8c41d12,
name: 'Giraia' } ],
members:
[ { _id: 601817dc2eeecd17db3a68f6,
name: 'Maria' },
{ _id: 601817ef2eeecd17db3a68f7,
name: 'Jose' },
{ _id: 60182cbb2b654330d2458f89,
name: 'Jonas'} ] }
管道的最后一步是比较 arrays,过滤哪些members
不是candidates
成员并将它们添加到数组stillAvailable
中。 我尝试了很多方法,但我无法通过聚合实现我的目标。 我能找到的唯一解决方案是在我的后端处理不完整管道的结果。 代码是:
orders.forEach(
(order) =>
(order.stillAvailable = order.members.filter(
(autonomous) =>
!order.candidate.some((el) => {
return el._id.toString() === autonomous._id.toString();
})
))
);
有了这个,我达到了预期的 output ......
{ candidate:
[ { _id: 601817dc2eeecd17db3a68f6,
name: 'Maria' },
{ _id: 601817ef2eeecd17db3a68f7,
name: 'Jose' } ],
groupData:
[ { _id: 606632403fffb851b8c41d12,
name: 'Giraia' ],
members:
[ { _id: 601817dc2eeecd17db3a68f6,
name: 'Maria' },
{ _id: 601817ef2eeecd17db3a68f7,
name: 'Jose' },
{ _id: 60182cbb2b654330d2458f89,
name: 'Jonas' ],
stillAvailable:
[ { _id: 60182cbb2b654330d2458f89,
name: 'Jonas' ] }
问题是为了更好地划分我的代码,有必要实现最后一步(在我的后端使用 javascript 完成)作为管道上的另一个步骤。 有谁知道如何达到这个目标?
在我在这里写下问题之后,不知何故,这个想法结构更好,我使用$map
和更高级别的$lookup
实现了结果。 我在这里留下了答案,以防有人遇到同样的问题。
let orders = await Order.aggregate(
{
$unwind: "$candidate",
},
{
$lookup: {
from: "groups",
localField: "candidate.groupId",
foreignField: "_id",
as: "groupData",
},
},
{
$lookup: {
from: "users",
let: {
id: "$candidate.groupId",
},
pipeline: [
{ $match: { groupId: { $ne: null } } },
{
$match: {
$expr: { $in: ["$$id", "$groupId"] },
},
},
{ $project: { name: 1, email: 1, _id: 1 } },
],
as: "members",
},
},
{ $match: { "members._id": new ObjectId(req.userId) } },
{
$lookup: {
from: "users",
let: { ids: "$candidate.autonomousId" },
pipeline: [
{
$match: {
$expr: { $in: ["$_id", "$$ids"] },
},
},
{ $project: { name: 1, email: 1, _id: 1 } },
],
as: "candidate",
},
},
{
$project: {
groupData: 1,
members: 1,
candidate: 1,
_id: 0,
stillAvailable: {
$setDifference: [
{
$map: {
input: "$members",
as: "member",
in: "$$member._id",
},
},
{
$map: {
input: "$candidate",
as: "el",
in: "$$el._id",
},
},
],
},
},
},
{
$lookup: {
from: "users",
let: {
ids: "$stillAvailable",
},
pipeline: [
{
$match: {
$expr: { $in: ["$_id", "$$ids"] },
},
},
{ $project: { name: 1, email: 1, _id: 1 } },
],
as: "stillAvailable",
},
}
).toArray();
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