如果 $match 条件总计不满足,如何将此 mongo 查询更改为 return "00:00"?
How to change this mongo query to return "00:00" if $match condition does not satisfy in aggregate?
假设我有一个这样的时间表集合:
[
{
_id: 1,
createdBy: "John",
duration: "00:30"
},
{
_id: 2,
createdBy: "John",
duration: "01:30"
},
{
_id: 3,
createdBy: "Mark",
duration: "00:30"
},
]
我的输入是一组用户名:
["John", "Mark", "Rikio"]
我想使用 mongodb 聚合来计算用户名数组中每个用户的时间表总持续时间,如果没有找到时间表,它应该 return duration: "00:00 ”。例如,它应该 return:
[
{createdBy: "John", totalDuration: "02:00"},
{createdBy: "Mark", totalDuration: "00:30"},
{createdBy: "Rikio", totalDuration: "00:00"}
]
但是,当我使用 $match 查询时,如果没有时间表,它不会 return 任何东西,所以我不知道哪个用户没有时间表和 return "00: 00" 给他们。
我完全同意@turivishal,但仍然可以通过 mongo 一个丑陋的查询来完成。
db.collection.aggregate([
{
$match: {}
},
{
$set: {
minutes: {
$let: {
vars: {
time: {
$split: [
"$duration",
":"
]
}
},
in: {
"$add": [
{
"$multiply": [
{
$toInt: {
$first: "$$time"
}
},
60
]
},
{
$toInt: {
$last: "$$time"
}
}
]
}
}
}
}
},
{
$group: {
"_id": "$createdBy",
"totalMinutes": {
"$sum": "$minutes"
}
}
},
{
$group: {
"_id": null,
"docs": {
"$push": "$$ROOT"
}
}
},
{
$set: {
"docs": {
$map: {
"input": [
"John",
"Mark",
"Rikio"
],
"as": "name",
"in": {
$let: {
vars: {
findName: {
"$first": {
"$filter": {
"input": "$docs",
"as": "d",
"cond": {
"$eq": [
"$$d._id",
"$$name"
]
}
}
}
}
},
in: {
"$cond": {
"if": "$$findName",
"then": "$$findName",
"else": {
_id: "$$name",
totalMinutes: 0
}
}
}
}
}
}
}
}
},
{
$unwind: "$docs"
},
{
$replaceRoot: {
"newRoot": "$docs"
}
},
{
$set: {
"hours": {
$floor: {
"$divide": [
"$totalMinutes",
60
]
}
},
"minutes": {
"$mod": [
"$totalMinutes",
60
]
}
}
},
{
$set: {
"hours": {
"$cond": {
"if": {
"$lt": [
"$hours",
10
]
},
"then": {
"$concat": [
"0",
{
"$toString": "$hours"
}
]
},
"else": {
"$toString": "$hours"
}
}
},
"minutes": {
"$cond": {
"if": {
"$lt": [
"$minutes",
10
]
},
"then": {
"$concat": [
"0",
{
"$toString": "$minutes"
}
]
},
"else": {
"$toString": "$minutes"
}
}
}
}
},
{
$project: {
duration: {
"$concat": [
"$hours",
":",
"$minutes"
]
}
}
}
])
假设我有一个这样的时间表集合:
[
{
_id: 1,
createdBy: "John",
duration: "00:30"
},
{
_id: 2,
createdBy: "John",
duration: "01:30"
},
{
_id: 3,
createdBy: "Mark",
duration: "00:30"
},
]
我的输入是一组用户名:
["John", "Mark", "Rikio"]
我想使用 mongodb 聚合来计算用户名数组中每个用户的时间表总持续时间,如果没有找到时间表,它应该 return duration: "00:00 ”。例如,它应该 return:
[
{createdBy: "John", totalDuration: "02:00"},
{createdBy: "Mark", totalDuration: "00:30"},
{createdBy: "Rikio", totalDuration: "00:00"}
]
但是,当我使用 $match 查询时,如果没有时间表,它不会 return 任何东西,所以我不知道哪个用户没有时间表和 return "00: 00" 给他们。
我完全同意@turivishal,但仍然可以通过 mongo 一个丑陋的查询来完成。
db.collection.aggregate([
{
$match: {}
},
{
$set: {
minutes: {
$let: {
vars: {
time: {
$split: [
"$duration",
":"
]
}
},
in: {
"$add": [
{
"$multiply": [
{
$toInt: {
$first: "$$time"
}
},
60
]
},
{
$toInt: {
$last: "$$time"
}
}
]
}
}
}
}
},
{
$group: {
"_id": "$createdBy",
"totalMinutes": {
"$sum": "$minutes"
}
}
},
{
$group: {
"_id": null,
"docs": {
"$push": "$$ROOT"
}
}
},
{
$set: {
"docs": {
$map: {
"input": [
"John",
"Mark",
"Rikio"
],
"as": "name",
"in": {
$let: {
vars: {
findName: {
"$first": {
"$filter": {
"input": "$docs",
"as": "d",
"cond": {
"$eq": [
"$$d._id",
"$$name"
]
}
}
}
}
},
in: {
"$cond": {
"if": "$$findName",
"then": "$$findName",
"else": {
_id: "$$name",
totalMinutes: 0
}
}
}
}
}
}
}
}
},
{
$unwind: "$docs"
},
{
$replaceRoot: {
"newRoot": "$docs"
}
},
{
$set: {
"hours": {
$floor: {
"$divide": [
"$totalMinutes",
60
]
}
},
"minutes": {
"$mod": [
"$totalMinutes",
60
]
}
}
},
{
$set: {
"hours": {
"$cond": {
"if": {
"$lt": [
"$hours",
10
]
},
"then": {
"$concat": [
"0",
{
"$toString": "$hours"
}
]
},
"else": {
"$toString": "$hours"
}
}
},
"minutes": {
"$cond": {
"if": {
"$lt": [
"$minutes",
10
]
},
"then": {
"$concat": [
"0",
{
"$toString": "$minutes"
}
]
},
"else": {
"$toString": "$minutes"
}
}
}
}
},
{
$project: {
duration: {
"$concat": [
"$hours",
":",
"$minutes"
]
}
}
}
])