时间:2021-07-01 10:21:17 帮助过:21人阅读
db.orders.aggregate([
{
$group:{
_id:null,
count:{$sum:1}
}
}
])

例2:
SQL:
SELECT SUM(price) AS total FROM orders
Mongodb:
db.orders.aggregate(
[
{
$group:
{
_id:null,
total:{$sum:"$price"}
}
}
])

例3:
SQL:
SELECT cust_id,SUM(price) AS total FROM orders GROUP BY cust_id
Mongodb:
db.orders.aggregate([
{
$group:
{
_id:"$cust_id",
total:
{
$sum:"$price"
}
}
},
{
$sort:
{
total:1
}
}
])

例4:
SQL:
SELECT cust_id, ord_date,SUM(price) AS total FROM orders GROUP BY cust_id, ord_date
Mongodb:
db.orders.aggregate([
{
$group:
{
_id:
{
cust_id:"$cust_id",
ord_date:
{
month:{$month:"$ord_date"},
day:{$dayOfMonth:"$ord_date"},
year:{$year:"$ord_date"}
}
},
total:{$sum:"$price"}
}
}
])

(译文)SQL与Mongodb聚合之前的对应关系
标签:match status price 参考 log group by 例子 聚合函数 ref