问题描述
说我有三个学生...
Lets say I have three students...
爱丽丝，她总是在星期五。
Alice, she is Always there on fridays.
{
"name" : "Alice",
"goes" : {
"mondays" : {
"fr" : 900,
"to" : 1400
},
"fridays" : {
"fr" : 700,
"to" : 1600
},
}
}
还有鲍勃，这里应该是第一个一月
And bob, here should be there on the first of january
{
"_id" : ObjectId("5284a7085d60338b40b8f17d"),
"name" : "Bob",
"goes" : {
"mondays" : {
"fr" : 800,
"to" : 1200
},
"special" : [
{
"date" : "2010-01-01",
"fr" : 1000,
"to" : 1500
}
]
}
}
和克莱尔谁不会周一参加r为10.00
And Clair who will not be attenging on mondays or at 10.00
{
"_id" : ObjectId("5284c2785d60338b40b8f17f"),
"name" : "Clair",
"goes" : {
"wednesdays" : {
"fr" : 1100,
"to" : 1500
},
"special" : [
{
"date" : "2010-01-01",
"fr" : 1600,
"to" : 1900
},
{
"date" : "2010-01-02",
"fr" : 1000,
"to" : 1300
}
]
}
}
我想找到所有应该
所以我在汇总框架中做到这一点。
So I do this with the aggregation framework.
db.students.aggregate(
[
{
$unwind: "$goes.special"
},
{
$match: {
$or : [
{
'goes.fridays.fr': 700,
},
{
'goes.special.date' : '2010-01-01',
'goes.special.fr': 1000
}
]
}
}
]
)
但是Alice没有出现。它明确指出了为什么在mongodb文档中在最底部。
But Alice does not show up. It clearly states why in the mongodb docs, http://docs.mongodb.org/manual/reference/operator/aggregation/unwind/ at the very bottom.
我可以通过添加一个带有空值的数组来解决它，但这并不像一个好的解决方案。
I could solve it by adding an array with a null value in it but that does not seam like a nice solution.
是否有一种方法可以让我放心，不要忽略在$ unwind'ed数组中没有数据的文档？
Is there a way I could get unwind NOT to ignore documents that does not have data in a $unwind'ed array?
推荐答案
您完全不需要 $ unwind 。简单的 $ match 在管道中就足够了：
You don't need $unwind at all. Simple $match in pipeline is enough:
pipeline = [
{
"$match" : {
"$or" : [
{
"goes.fridays.fr" : 700
},
{
"goes.special" : {
"$elemMatch" : {
"date" : "2010-01-01",
"fr" : 1000
}
}
}
]
}
}
]
db.students.aggregate(pipeline)
即使没有聚合框架也可以轻松实现。
It can be done easily even without aggregation framework.
query = {
"$or" : [
{
"goes.fridays.fr" : 700
},
{
"goes.special" : {
"$elemMatch" : {
"date" : "2010-01-01",
"fr" : 1000
}
}
}
]
}
db.students.find(query)
这篇关于如何使用$ unwind保持文档聚合的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！