【Mongo】聚合函数

http://blog.csdn.net/miyatang/article/details/20997313

 

SQL Terms, Functions, and Concepts
MongoDB Aggregation Operators
WHERE
$match
GROUP BY
$group
HAVING
$match
SELECT
$project
ORDER BY
$sort
LIMIT
$limit
SUM()
$sum
COUNT()
$sum
join
No direct corresponding operator; however, the $unwindoperator allows for somewhat similar functionality, but with fields embedded within the document.
实例：
[td]
SQL Example
MongoDB Example
Description
SELECT COUNT(*) AS countFROM orders

db.orders.aggregate( [ { $group: { _id: null, count: { $sum: 1 } } }] )

Count all records fromorders
SELECT SUM(price) AS totalFROM orders

db.orders.aggregate( [ { $group: { _id: null, total: { $sum: "$price" } } }] )

Sum theprice field from orders，这个非常有用，看官方说明，说_ID是必须，但没想到可以为NULL,
SELECT cust_id, SUM(price) AS totalFROM ordersGROUP BY cust_id

db.orders.aggregate( [ { $group: { _id: "$cust_id", total: { $sum: "$price" } } }] )

For each uniquecust_id, sum the pricefield.
SELECT cust_id, SUM(price) AS totalFROM ordersGROUP BY cust_idORDER BY total

db.orders.aggregate( [ { $group: { _id: "$cust_id", total: { $sum: "$price" } } }, { $sort: { total: 1 } }] )

For each uniquecust_id, sum the pricefield, results sorted by sum.
SELECT cust_id, ord_date, SUM(price) AS totalFROM ordersGROUP BY cust_id, ord_date

db.orders.aggregate( [ { $group: { _id: { cust_id: "$cust_id", ord_date: "$ord_date" }, total: { $sum: "$price" } } }] )

For each uniquecust_id,ord_dategrouping, sum the pricefield.
SELECT cust_id, count(*)FROM ordersGROUP BY cust_idHAVING count(*) > 1

db.orders.aggregate( [ { $group: { _id: "$cust_id", count: { $sum: 1 } } }, { $match: { count: { $gt: 1 } } }] )

For cust_idwith multiple records, return thecust_id and the corresponding record count.
SELECT cust_id, ord_date, SUM(price) AS totalFROM ordersGROUP BY cust_id, ord_dateHAVING total > 250

db.orders.aggregate( [ { $group: { _id: { cust_id: "$cust_id", ord_date: "$ord_date" }, total: { $sum: "$price" } } }, { $match: { total: { $gt: 250 } } }] )

For each uniquecust_id,ord_dategrouping, sum the pricefield and return only where the sum is greater than 250.
SELECT cust_id, SUM(price) as totalFROM ordersWHERE status = 'A'GROUP BY cust_id

db.orders.aggregate( [ { $match: { status: 'A' } }, { $group: { _id: "$cust_id", total: { $sum: "$price" } } }] )

For each uniquecust_id with status A, sum the pricefield.
SELECT cust_id, SUM(price) as totalFROM ordersWHERE status = 'A'GROUP BY cust_idHAVING total > 250

db.orders.aggregate( [ { $match: { status: 'A' } }, { $group: { _id: "$cust_id", total: { $sum: "$price" } } }, { $match: { total: { $gt: 250 } } }] )

For each uniquecust_id with status A, sum the pricefield and return only where the sum is greater than 250.
SELECT cust_id, SUM(li.qty) as qtyFROM orders o, order_lineitem liWHERE li.order_id = o.idGROUP BY cust_id

db.orders.aggregate( [ { $unwind: "$items" }, { $group: { _id: "$cust_id", qty: { $sum: "$items.qty" } } }] )

For each uniquecust_id, sum the corresponding line item qtyfields associated with the orders.
SELECT COUNT(*)FROM (SELECT cust_id, ord_date FROM orders GROUP BY cust_id, ord_date) as DerivedTable

db.orders.aggregate( [ { $group: { _id: { cust_id: "$cust_id", ord_date: "$ord_date" } } }, { $group: { _id: null, count: { $sum: 1 } } }] )