mongodb 在聚合框架中给出完整时间戳时如何按日期聚合？
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How to aggregate by date when a full timestamp is given in aggregation framework?
提问by BreakPhreak
I have a collection of errors, so that every error carries a datefield. How can I aggregate/count/group the errors by DAY only (i.e. exclude the time of the day)? I guess, some smart projection should be applied.
我有一系列错误，因此每个错误都带有一个date字段。如何仅按天汇总/计数/分组错误（即排除一天中的时间）？我想，应该应用一些智能投影。
回答by Philipp
You can do this by using the following aggregation operators:
您可以使用以下聚合运算符来执行此操作：
This gives you the error count for each date:
这为您提供了每个日期的错误计数：
db.errors.aggregate(
{ $group : {
_id: {
year : { $year : "$date" },
month : { $month : "$date" },
day : { $dayOfMonth : "$date" },
},
count: { $sum: 1 }
}}
);
This example assumes that the date field in your error documents is dateand of type BSON Date. There is also a Timestamp type in MongoDB, but use of this type is explicitely discouraged by the documentation:
此示例假定您的错误文档中的日期字段是BSON Datedate类型。MongoDB 中还有一个 Timestamp 类型，但文档明确不鼓励使用这种类型：
Note: The BSON Timestamp type is for internal MongoDB use. For most cases, in application development, you will want to use the BSON date type. See Date for more information.
注意：BSON 时间戳类型供内部 MongoDB 使用。大多数情况下，在应用程序开发中，您会希望使用 BSON 日期类型。有关详细信息，请参阅日期。
回答by Dmitry Sergeev
You can convert your timestamp field to string with specific date format by using $project (aggregation)
您可以使用$project (aggregation)将时间戳字段转换为具有特定日期格式的字符串
aggregate([
{
'$project': {
newFieldName: {'$dateToString': {format: '%Y-%m-%d', date: '$yourDateFieldName'}}
}
}, {
'$group': {
_id: {newFieldName: '$newFieldName'},
viewCount: {'$sum': 1}
}
},
{'$sort': {'_id.newFieldName': 1}}
], {})
