MongoDB中是否有计算字段?
在SQL中,我可以这样做:
SELECT A+B AS C FROM MYTABLE WHERE C>10
我可以在MongoDB中做类似的事情吗?
更新
我做了投影:
db.segments.aggregate(
[
{
$project: {
"_id": 1,
numberOfRestrictions: { $size: "$Speed Restrictions" }
}
}
]
)
它有效。
不幸的是,进一步的流水线操作不会:
db.segments.aggregate(
[
{
$project: {
"_id": 1,
numberOfRestrictions: { $size: "$Speed Restrictions" }
}
},
{
$match: {
"numberOfRestrictions": {
"$gt": 1
}
}
}
]
)
后期导致错误
The argument to $size must be an Array, but was of type: EOO
答案 0 :(得分:5)
是。它被称为aggregation pipelines。具体而言,您需要使用 $project
阶段来创建 C
字段,然后使用 $match
< / strong>阶段查找符合条件的所有文件。
让我们先创建一些文档:
for( var i = 1; i <=10; i++){
db.agg.insert({a:i,b:i})
}
这导致集合看起来像这样:
> db.agg.find()
{ "_id" : ObjectId("56c1b5561a3b578f37a99d4d"), "a" : 1, "b" : 1 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d4e"), "a" : 2, "b" : 2 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d4f"), "a" : 3, "b" : 3 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d50"), "a" : 4, "b" : 4 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d51"), "a" : 5, "b" : 5 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d52"), "a" : 6, "b" : 6 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d53"), "a" : 7, "b" : 7 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d54"), "a" : 8, "b" : 8 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d55"), "a" : 9, "b" : 9 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d56"), "a" : 10, "b" : 10 }
db.agg.aggregate([
// You need to include all fields you want to have
// in the resulting document within the $project stage
{ "$project":{ a:1, b:1, c:{ "$add": ["$a","$b"] }}},
{ "$match":{ c:{ "$gt":10 }}}
])
返回以下结果:
{ "_id" : ObjectId("56c1b5561a3b578f37a99d52"), "a" : 6, "b" : 6, "c" : 12 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d53"), "a" : 7, "b" : 7, "c" : 14 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d54"), "a" : 8, "b" : 8, "c" : 16 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d55"), "a" : 9, "b" : 9, "c" : 18 }
{ "_id" : ObjectId("56c1b5561a3b578f37a99d56"), "a" : 10, "b" : 10, "c" : 20 }
答案 1 :(得分:4)