mongoDB与sql聚合操作对应图

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.

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) AStotalFROM 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) AStotalFROM ordersGROUP BYcust_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 ordersGROUPBY 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(*)FROMordersGROUP 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 ordersGROUPBY 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) astotalFROM 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) astotalFROM 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) asqtyFROM orders o,     order_lineitem liWHERE li.order_id = o.idGROUP BYcust_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 BYcust_id, ord_date) as DerivedTable

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