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  • 【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 } } }] )

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  • 原文地址:https://www.cnblogs.com/colipso/p/4560495.html
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