在 MongoDB 中按条件分组
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Group By Condition in MongoDB
提问by okoboko
I have a series of documents (check events) in MongoDB that look like this:
我在 MongoDB 中有一系列文件（检查事件），如下所示：
{
"_id" : ObjectId("5397a78ab87523acb46f56"),
"inspector_id" : ObjectId("5397997a02b8751dc5a5e8b1"),
"status" : 'defect',
"utc_timestamp" : ISODate("2014-06-11T00:49:14.109Z")
}
{
"_id" : ObjectId("5397a78ab87523acb46f57"),
"inspector_id" : ObjectId("5397997a02b8751dc5a5e8b2"),
"status" : 'ok',
"utc_timestamp" : ISODate("2014-06-11T00:49:14.109Z")
}
I need to get a result set that looks like this:
我需要得到一个如下所示的结果集：
[
{
"date" : "2014-06-11",
"defect_rate" : '.92'
},
{
"date" : "2014-06-11",
"defect_rate" : '.84'
},
]
In other words, I need to get the average defect rate per day. Is this possible?
换句话说，我需要获得每天的平均缺陷率。这可能吗？
回答by Neil Lunn
The aggregation framework is what you want:
聚合框架就是你想要的：
db.collection.aggregate([
{ "$group": {
"_id": {
"year": { "$year": "$utc_timestamp" },
"month": { "$month": "$utc_timestamp" },
"day": { "$dayOfMonth": "$utc_timestamp" },
},
"defects": {
"$sum": { "$cond": [
{ "$eq": [ "$status", "defect" ] },
1,
0
]}
},
"totalCount": { "$sum": 1 }
}},
{ "$project": {
"defect_rate": {
"$cond": [
{ "$eq": [ "$defects", 0 ] },
0,
{ "$divide": [ "$defects", "$totalCount" ] }
]
}
}}
])
So first you group on the day using the date aggregation operatorsand get the totalCount of items on the given day. The use of the $condoperator here determines whether the "status" is actually a defect or not and the result is a conditional $sumwhere only the "defect" values are counted.
因此，首先您使用日期聚合运算符对当天进行分组，并获取给定日期的项目总数。$cond此处使用运算符确定“状态”是否实际上是缺陷，结果是$sum仅计算“缺陷”值的条件。
Once those are grouped per day you simply $dividethe result, with another check with $condto make sure you are not dividing by zero.
