nir*_*air 6 date mongodb mongodb-query aggregation-framework
这是查询
[
{
"$project": {
"formattedDate": {
"$dateToString": { "format": "%Y-%m-%d", "date": "$ceatedAt" }
},
"createdAtMonth": { "$month": "$ceatedAt" },
"rating": 1
}
},
{
"$group": {
"_id": "$formattedDate",
"average": { "$avg": "$rating" },
"month": { "$first": "$createdAtMonth" },
}
}
]
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我需要时间戳中的日期.怎么做?
tar*_*pka 11
使用算术集成算你的日期作为被减数,并作为减数.$subtract
new Date("1970-01-01")
db.collection.aggregate(
{
$project: { "timestamp": { $subtract: [ "$createdAt", new Date("1970-01-01") ] } }
}
);
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对于文件
{ "_id": 1, "createdAt": ISODate("2016-09-01T14:35:14.952Z") }
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结果是
{ "_id": 1, "timestamp": NumberLong("1472740514952") }
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如果要按时间戳和(年,月,日)对两者进行分组,则可以将时间戳除以一天中的毫秒数,以使其在每一天(而不是每毫秒)都是唯一的
db.collection.aggregate(
{
$project:
{
"timestampByDay":
{
$floor:
{
$divide:
[
{ $subtract: [ "$createdAt", new Date("1970-01-01") ] },
24 * 60 * 60 * 1000
]
}
},
"date": "$createdAt"
}
},
{
$group:
{
"_id": "$timestampByDay",
"date": { $first: "$date" }
}
}
);
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Ash*_*shh 10
Mongodb 4.0引入了$toLong
将日期转换为时间戳的聚合
db.collection.aggregate([
{ "$project": {
"createdAt": {
"$toLong": "$createdAt"
}
}}
])
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你可以在这里试试