我有以下查询：
db.getCollection('user').aggregate([
{$unwind: "$education"},
{$project: {
duration: {"$divide":[{$subtract: ['$education.to', '$education.from'] }, 1000 * 60 * 60 * 24 * 365]}
}},
{$group: {
_id: '$_id',
"duration": {$sum: '$duration'}
}}]
])
以上查询结果为：
{
"_id" : ObjectId("59fabb20d7905ef056f55ac1"),
"duration" : 2.34794520547945
}
/* 2 */
{
"_id" : ObjectId("59fab630203f02f035301fc3"),
"duration" : 2.51232876712329
}
但我想要做的是以year + month + day格式获取其持续时间，例如：2 y, 3 m, 20 d。
另外一点，如果to字段上的课程为空，另一个字段为isGoingOn: true，那么我应该使用当前日期而不是to字段来计算持续时间。
用户有一系列课程子文档
education: [
{
"courseName": "Java",
"from" : ISODate("2010-12-08T00:00:00.000Z"),
"to" : ISODate("2011-05-31T00:00:00.000Z"),
"isGoingOn": false
},
{
"courseName": "PHP",
"from" : ISODate("2013-12-08T00:00:00.000Z"),
"to" : ISODate("2015-05-31T00:00:00.000Z"),
"isGoingOn": false
},
{
"courseName": "Mysql",
"from" : ISODate("2017-02-08T00:00:00.000Z"),
"to" : null,
"isGoingOn": true
}
]
另一点是：该日期在一个子文档中可能不连续到另一个子文档。用户可能有1年的课程，然后在两年后，他/她开始他/她的下一个课程1年和3个月（这意味着该用户总共有2年和3个月的课程持续时间） 。
我想要的是在educations数组中获取每个子文档的日期差异，并将它们相加。假设我的样本数据Java课程持续时间为6个月，22天，PHP课程持续时间为1年，6个月和22天，最后一个课程为2017年2月8日至今，它正在发生，所以我的教育持续时间是这些间隔的总和。
答案 0 :(得分：3)
请尝试此聚合以获取日期，月份和年份的日期差异，添加多个$addFields阶段计算并减少日期差异，月份范围不下溢，此处假设为1个月= 30天
管道
db.edu.aggregate(
[
{
$addFields : {
trainingPeriod : {
$map : {
input : "$education",
as : "t",
in : {
year: {$subtract: [{$year : {$ifNull : ["$$t.to", new Date()]}}, {$year : "$$t.from"}]},
month: {$subtract: [{$month : {$ifNull : ["$$t.to", new Date()]}}, {$month : "$$t.from"}]},
dayOfMonth: {$subtract: [{$dayOfMonth : {$ifNull : ["$$t.to", new Date()]}}, {$dayOfMonth : "$$t.from"}]}
}
}
}
}
},
{
$addFields : {
trainingPeriod : {
$map : {
input : "$trainingPeriod",
as : "d",
in : {
year: "$$d.year",
month: {$cond : [{$lt : ["$$d.dayOfMonth", 0]}, {$subtract : ["$$d.month", 1]}, "$$d.month" ]},
day: {$cond : [{$lt : ["$$d.dayOfMonth", 0]}, {$add : [30, "$$d.dayOfMonth"]}, "$$d.dayOfMonth" ]}
}
}
}
}
},
{
$addFields : {
trainingPeriod : {
$map : {
input : "$trainingPeriod",
as : "d",
in : {
year: {$cond : [{$lt : ["$$d.month", 0]}, {$subtract : ["$$d.year", 1]}, "$$d.year" ]},
month: {$cond : [{$lt : ["$$d.month", 0]}, {$add : [12, "$$d.month"]}, "$$d.month" ]},
day: "$$d.day"
}
}
}
}
},
{
$addFields : {
total : {
$reduce : {
input : "$trainingPeriod",
initialValue : {year : 0, month : 0, day : 0},
in : {
year: {$add : ["$$this.year", "$$value.year"]},
month: {$add : ["$$this.month", "$$value.month"]},
day: {$add : ["$$this.day", "$$value.day"]}
}
}
}
}
},
{
$addFields : {
total : {
year : "$total.year",
month : {$add : ["$total.month", {$floor : {$divide : ["$total.day", 30]}}]},
day : {$mod : ["$total.day", 30]}
}
}
},
{
$addFields : {
total : {
year : {$add : ["$total.year", {$floor : {$divide : ["$total.month", 12]}}]},
month : {$mod : ["$total.month", 12]},
day : "$total.day"
}
}
}
]
).pretty()
结果
{
"_id" : ObjectId("5a895d4721cbd77dfe857f95"),
"education" : [
{
"courseName" : "Java",
"from" : ISODate("2010-12-08T00:00:00Z"),
"to" : ISODate("2011-05-31T00:00:00Z"),
"isGoingOn" : false
},
{
"courseName" : "PHP",
"from" : ISODate("2013-12-08T00:00:00Z"),
"to" : ISODate("2015-05-31T00:00:00Z"),
"isGoingOn" : false
},
{
"courseName" : "Mysql",
"from" : ISODate("2017-02-08T00:00:00Z"),
"to" : null,
"isGoingOn" : true
}
],
"trainingPeriod" : [
{
"year" : 0,
"month" : 5,
"day" : 23
},
{
"year" : 1,
"month" : 5,
"day" : 23
},
{
"year" : 1,
"month" : 0,
"day" : 10
}
],
"total" : {
"year" : 2,
"month" : 11,
"day" : 26
}
}
>
答案 1 :(得分：0)
那么你可以简单地使用现有的date aggregation operator s而不是使用数学转换为＆＃34; days＆＃34;你现在有：
db.getCollection('user').aggregate([
{ "$unwind": "$education" },
{ "$group": {
"_id": "$_id",
"years": {
"$sum": {
"$subtract": [
{ "$subtract": [
{ "$year": { "$ifNull": [ "$education.to", new Date() ] } },
{ "$year": "$education.from" }
]},
{ "$cond": {
"if": {
"$gt": [
{ "$month": { "$ifNull": [ "$education.to", new Date() ] } },
{ "$month": "$education.from" }
]
},
"then": 0,
"else": 1
}}
]
}
},
"months": {
"$sum": {
"$add": [
{ "$subtract": [
{ "$month": { "$ifNull": [ "$education.to", new Date() ] } },
{ "$month": "$education.from" }
]},
{ "$cond": {
"if": {
"$gt": [
{ "$month": { "$ifNull": ["$education.to", new Date() ] } },
{ "$month": "$education.from" }
]
},
"then": 0,
"else": 12
}}
]
}
},
"days": {
"$sum": {
"$add": [
{ "$subtract": [
{ "$dayOfYear": { "$ifNull": [ "$education.to", new Date() ] } },
{ "$dayOfYear": "$education.from" }
]},
{ "$cond": {
"if": {
"$gt": [
{ "$month": { "$ifNull": [ "$education.to", new Date() ] } },
{ "$month": "$education.from" }
]
},
"then": 0,
"else": 365
}}
]
}
}
}},
{ "$project": {
"years": {
"$add": [
"$years",
{ "$add": [
{ "$floor": { "$divide": [ "$months", 12 ] } },
{ "$floor": { "$divide": [ "$days", 365 ] } }
]}
]
},
"months": {
"$mod": [
{ "$add": [
"$months",
{ "$floor": {
"$multiply": [
{ "$divide": [ "$days", 365 ] },
12
]
}}
]},
12
]
},
"days": { "$mod": [ "$days", 365 ] }
}}
])
它是＆＃34;有点＆＃34;在＆＃34;天＆＃34;的近似值和＆＃34;月＆＃34;没有必要的操作＆＃34;某些＆＃34;闰年，但它会得到你应该＆＃34;足够接近＆＃34;用于大多数目的。
只要您的MongoDB版本为3.2或更高版本，您甚至可以在没有$unwind的情况下执行此操作：
db.getCollection('user').aggregate([
{ "$addFields": {
"duration": {
"$let": {
"vars": {
"edu": {
"$map": {
"input": "$education",
"as": "e",
"in": {
"$let": {
"vars": { "toDate": { "$ifNull": ["$$e.to", new Date()] } },
"in": {
"years": {
"$subtract": [
{ "$subtract": [
{ "$year": "$$toDate" },
{ "$year": "$$e.from" }
]},
{ "$cond": {
"if": { "$gt": [{ "$month": "$$toDate" },{ "$month": "$$e.from" }] },
"then": 0,
"else": 1
}}
]
},
"months": {
"$add": [
{ "$subtract": [
{ "$ifNull": [{ "$month": "$$toDate" }, new Date() ] },
{ "$month": "$$e.from" }
]},
{ "$cond": {
"if": { "$gt": [{ "$month": "$$toDate" },{ "$month": "$$e.from" }] },
"then": 0,
"else": 12
}}
]
},
"days": {
"$add": [
{ "$subtract": [
{ "$ifNull": [{ "$dayOfYear": "$$toDate" }, new Date() ] },
{ "$dayOfYear": "$$e.from" }
]},
{ "$cond": {
"if": { "$gt": [{ "$month": "$$toDate" },{ "$month": "$$e.from" }] },
"then": 0,
"else": 365
}}
]
}
}
}
}
}
}
},
"in": {
"$let": {
"vars": {
"years": { "$sum": "$$edu.years" },
"months": { "$sum": "$$edu.months" },
"days": { "$sum": "$$edu.days" }
},
"in": {
"years": {
"$add": [
"$$years",
{ "$add": [
{ "$floor": { "$divide": [ "$$months", 12 ] } },
{ "$floor": { "$divide": [ "$$days", 365 ] } }
]}
]
},
"months": {
"$mod": [
{ "$add": [
"$$months",
{ "$floor": {
"$multiply": [
{ "$divide": [ "$$days", 365 ] },
12
]
}}
]},
12
]
},
"days": { "$mod": [ "$$days", 365 ] }
}
}
}
}
}
}}
])
这是因为从MongoDB 3.4开始，您可以$sum直接使用$addFields或$project等阶段的数组或任何表达式列表，$map可以应用那些相同的＆＃34;日期聚合运算符＆＃34;对每个数组元素的表达式代替首先执行$unwind。
所以主要的数学运算真的可以在＆＃34;减少＆＃34;的一部分完成。数组，然后每个总数可以由一般＆＃34;除数＆＃34;多年来，＆＃34; modulo＆＃34;或＆＃34;余数＆＃34;来自几个月和几天的任何超支。
基本上返回：
{
"_id" : ObjectId("5a07688e98e4471d8aa87940"),
"education" : [
{
"courseName" : "Java",
"from" : ISODate("2010-12-08T00:00:00.000Z"),
"to" : ISODate("2011-05-31T00:00:00.000Z"),
"isGoingOn" : false
},
{
"courseName" : "PHP",
"from" : ISODate("2013-12-08T00:00:00.000Z"),
"to" : ISODate("2015-05-31T00:00:00.000Z"),
"isGoingOn" : false
},
{
"courseName" : "Mysql",
"from" : ISODate("2017-02-08T00:00:00.000Z"),
"to" : null,
"isGoingOn" : true
}
],
"duration" : {
"years" : 3.0,
"months" : 3.0,
"days" : 259.0
}
}
鉴于2017年11月11日
答案 2 :(得分：0)
您可以使用moment js库进行客户端处理来简化代码。
所有日期时间数学都由时刻js库处理。使用duration计算缩短的时间diff
使用reduce在所有数组元素之间添加时间差异，然后按片段持续时间输出以年/月/天为单位的时间。
它解决了两个问题：
例如：
var education = [
{
"courseName": "Java",
"from" : new Date("2010-12-08T00:00:00.000Z"),
"to" : new Date("2011-05-31T00:00:00.000Z"),
"isGoingOn": false
},
{
"courseName": "PHP",
"from" : new Date("2013-12-08T00:00:00.000Z"),
"to" : new Date("2015-05-31T00:00:00.000Z"),
"isGoingOn": false
},
{
"courseName": "Mysql",
"from" : new Date("2017-02-08T00:00:00.000Z"),
"to" : null,
"isGoingOn": true
}
];
var reducedDiff = education.reduce(function(prevVal, elem) {
if(elem.isGoingOn) elem.to = new Date();
var diffDuration = moment(elem.to).diff(moment(elem.from));
return prevVal + diffDuration;
}, 0);
var duration = moment.duration(reducedDiff);
alert(duration.years() +" y, " + duration.months() + " m, " + duration.days() + " d " );
var durationstr = duration.years() +" y, " + duration.months() + " m, " + duration.days() + " d ";
MongoDb整合：
var reducedDiff = db.getCollection('user').find({},{education:1}).reduce(function(...