我正在尝试编写一个聚合来计算有多少文档具有某些字段(即只计算它们是否存在)。对象看起来像这样:
{
"_id" : ObjectId("5617731fe65e0b19101c7039"),
"dateCreated" : ISODate("2015-10-09T07:56:15.068Z"),
"dateSent" : ISODate("2015-10-09T07:56:16.682Z"),
"dateAttempted" : ISODate("2015-10-09T07:56:16.682Z")
},
{
"_id" : ObjectId("561e37bb537d381bb0ef0ae2"),
"dateCreated" : ISODate("2015-10-14T11:08:43.306Z"),
"dateSent" : ISODate("2015-10-14T11:09:51.618Z"),
"dateAttempted" : ISODate("2015-10-14T11:09:51.618Z"),
"dateViewed" : ISODate("2015-10-15T10:09:50.618Z"),
"dateOpened" : ISODate("2015-10-15T10:10:01.618Z")
}
我想遍历所有文档,计算该字段存在的位置。期望的输出:
{
"total" : 1000,
"created" : 1000,
"sent" : 990,
"attempted" : 995
"viewed" : 800,
"opened" : 750
}
如果每天可以将此输出分组,则可加分!我不希望对范围内的每个日期执行新的聚合。
这是我目前所拥有的,但没有用;它为每个字段返回零
[
{
"$group": {
"_id": {
"$dayOfMonth": "$dateCreated"
},
"total": {
"$sum": 1
},
"sent": {
"$sum": "$dateSent"
},
"attempted": {
"$sum": "$dateAttempted"
},
"viewed": {
"$sum": "$dateViewed"
},
"clicked": {
"$sum": "$dateClicked"
}
}
}
]
最佳答案
$cond
和 $ifNull
运算符是这里的 helper :
[
{
"$group": {
"_id": {
"$dayOfMonth": "$dateCreated"
},
"total": {
"$sum": 1
},
"sent": {
"$sum": { "$cond": [ { "$ifNull": [ "$dateSent", false ] }, 1, 0 ] }
},
"attempted": {
"$sum": { "$cond": [ { "$ifNull": [ "$dateAttempted", false ] }, 1, 0 ] }
},
"viewed": {
"$sum": { "$cond": [ { "$ifNull": [ "$dateViewed", false ] }, 1, 0 ] }
},
"clicked": {
"$sum": { "$cond": [ { "$ifNull": [ "$dateClicked", false ] }, 1, 0 ] }
}
}
}
]
$ifNull
将返回存在的字段(逻辑 true
)或替代值 false
。 $cond
查看此条件并返回 1
where true
或 0
where false 以提供条件数数。
关于mongodb - 当前字段的聚合条件计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33143156/