我有一个拥有50万条记录的员工集合。每条记录将具有以下详细信息。
mongo文档如下。
{
"_id": "234463456453643563456",
"name": "Mike",
"empId": "10",
"managerId": "8",
"projects" : [ "123", "456", "789"]
}
a。筛选 位置
b。过滤项目
结果应该像
10 ->>> Manager
/\
/ \
8 6 ---->> 8 & 6 reporting to manager 10
/\ /\
/ \ / \
4 5 2 1 ---->> 4 & 5 reporting to manager 8 ...
任何帮助您获得具有等级的分层结果的人都会感激吗？
我无法获得预期的结果。
样本数据：-
db.getCollection("employees").insert({"_id":"10","empId": "10","name":"Employee10","managerId":"15" });
db.getCollection("employees").insert({"_id":"8","empId": "8","name":"Employee8","managerId":"10" });
db.getCollection("employees").insert({"_id":"6","empId": "6","name":"Employee6","managerId":"10" });
db.getCollection("employees").insert({"_id":"4","empId": "4","name":"Employee4","managerId":"8" });
db.getCollection("employees").insert({"_id":"5","empId": "5","name":"Employee5","managerId":"8" });
db.getCollection("employees").insert({"_id":"2","empId": "2","name":"Employee2","managerId":"6" });
db.getCollection("employees").insert({"_id":"1","empId": "1","name":"Employee1","managerId":"6" });
查询：-
db.getCollection('employees').aggregate([
{
$match: {
empId : "10"
}
},
{
$graphLookup: {
from: "employees",
startWith: "$empId",
connectFromField: "empId",
connectToField: "managerId",
as: "reportees",
maxDepth: 4,
depthField: "level"
}
},
{
$project: {
"empId":1,
"managerId":1,
"reportees.empId":1,
"reportees.name":1,
"reportees.managerId":1,
"reportees.level":1
}
}
]);
实际结果：-
{
"_id" : "10",
"empId" : "10",
"managerId" : "15",
"reportees" : [
{
"empId" : "1",
"name" : "Employee1",
"managerId" : "6",
"level" : NumberLong(1)
},
{
"empId" : "4",
"name" : "Employee4",
"managerId" : "8",
"level" : NumberLong(1)
},
{
"empId" : "2",
"name" : "Employee2",
"managerId" : "6",
"level" : NumberLong(1)
},
{
"empId" : "5",
"name" : "Employee5",
"managerId" : "8",
"level" : NumberLong(1)
},
{
"empId" : "6",
"name" : "Employee6",
"managerId" : "10",
"level" : NumberLong(0)
},
{
"empId" : "8",
"name" : "Employee8",
"managerId" : "10",
"level" : NumberLong(0)
}
]
}
预期结果：-
{
"_id" : "10",
"empId" : "10",
"managerId" : "15",
"reportees" : [
{
"empId" : "6",
"name" : "Employee6",
"managerId" : "10",
"level" : NumberLong(0),
"reportees" : [
{
"empId" : "1",
"name" : "Employee1",
"managerId" : "6",
"level" : NumberLong(1)
},
{
"empId" : "2",
"name" : "Employee2",
"managerId" : "6",
"level" : NumberLong(1)
}
]
},
{
"empId" : "8",
"name" : "Employee8",
"managerId" : "10",
"level" : NumberLong(0),
"reportees" : [
{
"empId" : "5",
"name" : "Employee5",
"managerId" : "8",
"level" : NumberLong(1)
},
{
"empId" : "4",
"name" : "Employee4",
"managerId" : "8",
"level" : NumberLong(1)
}
]
}
]
}
问题：-
答案 0 :(得分：3)
这恰好是您$graphLookup所要达到的目标（至少需要遍历）。对于过滤部分，您可以根据要过滤的精确程度简单地使用$filter或$match。
看看这个查询的结果：
db.employees.aggregate({
$graphLookup: {
from: "employees",
startWith: "$managerId",
connectFromField: "managerId",
connectToField: "empId",
as: "managers",
}
})
根据您的澄清更新1：
为了获得您想要的层次结构，可以执行以下操作。但是，我不会将其称为一个漂亮的解决方案，因为它需要您静态定义要下降的级别数并重复部分，但这确实可以完成您的示例。不确定，是否可以轻松扩展到更多级别。我个人认为，客户端循环解决方案将更适合此类工作：
db.employees.aggregate([
{
$match: {
empId : "10"
}
},
// level 0
{
$graphLookup: {
from: "employees",
startWith: "$empId",
connectFromField: "empId",
connectToField: "managerId",
as: "reportees",
maxDepth: 0
}
},
{
$unwind: "$reportees" // flatten
},
{
$addFields: {
"reportees.level": 0 // add level field
}
},
// level 1
{
$graphLookup: {
from: "employees",
startWith: "$reportees.empId",
connectFromField: "reportees.empId",
connectToField: "managerId",
as: "reportees.reportees",
maxDepth: 0
}
},
{
$group: { // group previously flattened documents back together
_id: "$_id",
empId: { $first: "$empId" },
name: { $first: "$name" },
managerId: { $first: "$managerId" },
reportees: { $push: "$reportees" },
}
},
{
$addFields: {
"reportees.reportees.level": 1 // add level field
}
}
])
更新2：
以下查询从输出结构的角度将您带到您想要的位置（我省略了level字段，但是应该很容易添加）。但是，它并不是特别漂亮，并且再次要求您预先定义最大的组织深度。
db.employees.aggregate([
{
$match: {
empId : "10"
}
},
{
$graphLookup: { // get the relevant documents out of our universe of employees
from: "employees",
startWith: "$empId",
connectFromField: "empId",
connectToField: "managerId",
as: "reportees"
}
},
{
$project: { // add the employee we are interested in into the array of employees we're looking at
_id: 0,
reportees: { $concatArrays: [ "$reportees", [ { _id: "$_id", empId: "$empId", name: "$name", managerId: "$managerId" } ] ] }
}
},
{
$project: {
reportees: {
$let: {
vars: {
managers: {
$filter: { // remove employees with no reportess so keep managers only
input: {
$map: {
input: "$reportees",
as: "this",
in: {
$mergeObjects: [
"$$this",
{
reportees: {
$filter: { // extract reportees from list of employees
input: "$reportees",
as: "that",
cond: {
$eq: [ "$$this._id", "$$that.managerId" ]
}
}
}
}
]
}
}
},
as: "this",
cond: { $ne: [ "$$this.reportees", [] ] }
}
}
},
in: {
$cond: [ // this is to break the processing once we have reached a top level manager
{ $eq: [ "$$managers", [] ] },
"$reportees",
"$$managers"
]
}
}
}
}
},
// second level: exactly identical to the previous stage
// third level: exactly identical to the previous stage
// basically, from here onwards you would need to repeat an exact copy of the previous stage to go one level deeper
]);
答案 1 :(得分：3)
我相信拥有 level 字段，我们可以使用$reduce从数组构建层次结构。为此，我们需要按reportees之后的级别降序排列$graphLookup。不幸的是，目前唯一的方法是使用$unwind + $sort + $group，这会使聚合过程变得很长。
然后，我们可以使用$reduce处理该有序数组。在每个步骤中，我们只需要向结果集中添加一个雇员，包括上一级的reportees。另外，我们需要检测在处理过程中level何时发生变化，并在这种情况下重新排列辅助数组。
$addFields仅替换现有的reportees字段。 $concatArrays允许我们将当前员工（$$this）附加到结果中。使用$filter，我们可以从较低级别获得reportees。
db.getCollection('employees').aggregate([
{
$match: {
empId : "10"
}
},
{
$graphLookup: {
from: "employees",
startWith: "$empId",
connectFromField: "empId",
connectToField: "managerId",
as: "reportees",
maxDepth: 4,
depthField: "level"
}
},
{
$project: {
"empId":1,
"managerId":1,
"reportees.empId":1,
"reportees.name":1,
"reportees.managerId":1,
"reportees.level":1
}
},
{
$unwind: "$reportees"
},
{
$sort: { "reportees.level": -1 }
},
{
$group: {
_id: "$_id",
empId: { $first: "$empId" },
managerId: { $first: "$managerId" },
reportees: { $push: "$reportees" }
}
},
{
$addFields: {
reportees: {
$reduce: {
input: "$reportees",
initialValue: {
currentLevel: -1,
currentLevelEmployees: [],
previousLevelEmployees: []
},
in: {
$let: {
vars: {
prev: {
$cond: [
{ $eq: [ "$$value.currentLevel", "$$this.level" ] },
"$$value.previousLevelEmployees",
"$$value.currentLevelEmployees"
]
},
current: {
$cond: [
{ $eq: [ "$$value.currentLevel", "$$this.level" ] },
"$$value.currentLevelEmployees",
[]
]
}
},
in: {
currentLevel: "$$this.level",
previousLevelEmployees: "$$prev",
currentLevelEmployees: {
$concatArrays: [
"$$current",
[
{ $mergeObjects: [
"$$this",
{ reportees: { $filter: { input: "$$prev", as: "e", cond: { $eq: [ "$$e.managerId", "$$this.empId" ] } } } }
] }
]
]
}
}
}
}
}
}
}
},
{
$addFields: { reportees: "$reportees.currentLevelEmployees" }
}
]).pretty()
上述解决方案应适用于多个级别。输出：
{
"_id" : "10",
"empId" : "10",
"managerId" : "15",
"reportees" : [
{
"empId" : "6",
"name" : "Employee6",
"managerId" : "10",
"level" : NumberLong(0),
"reportees" : [
{
"empId" : "1",
"name" : "Employee1",
"managerId" : "6",
"level" : NumberLong(1),
"reportees" : [ ]
},
{
"empId" : "2",
"name" : "Employee2",
"managerId" : "6",
"level" : NumberLong(1),
"reportees" : [ ]
}
]
},
{
"empId" : "8",
"name" : "Employee8",
"managerId" : "10",
"level" : NumberLong(0),
"reportees" : [
{
"empId" : "5",
"name" : "Employee5",
"managerId" : "8",
"level" : NumberLong(1),
"reportees" : [ ]
},
{
"empId" : "4",
"name" : "Employee4",
"managerId" : "8",
"level" : NumberLong(1),
"reportees" : [ ]
}
]
}
]
}
答案 2 :(得分：0)
$ graphLookup上的官方文档或多或少会提供帮助。
https://docs.mongodb.com/manual/reference/operator/aggregation/graphLookup/
只是一种提醒。