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[MongoDB] SQL vs Aggregation Framework

데브포유 2017. 3. 21. 23:10
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SQL Terms, Functions, and ConceptsMongoDB Aggregation Operators
WHERE$match
GROUP BY$group
HAVING$match
SELECT$project
ORDER BY$sort
LIMIT$limit
SUM()$sum
COUNT()$sum
join

$lookup

New in version 3.2.

Examples

The following table presents a quick reference of SQL aggregation statements and the corresponding MongoDB statements. The examples in the table assume the following conditions:

  • The SQL examples assume two tables, orders and order_lineitem that join by the order_lineitem.order_id and the orders.id columns.

  • The MongoDB examples assume one collection orders that contain documents of the following prototype:

    {
  cust_id: "abc123",
  ord_date: ISODate("2012-11-02T17:04:11.102Z"),
  status: 'A',
  price: 50,
  items: [ { sku: "xxx", qty: 25, price: 1 },
           { sku: "yyy", qty: 25, price: 1 } ]
}
SQL ExampleMongoDB ExampleDescription
SELECT COUNT(*) AS count
FROM orders

db.orders.aggregate( [
   {
     $group: {
        _id: null,
        count: { $sum: 1 }
     }
   }
] )
Count all records from orders
SELECT SUM(price) AS total
FROM orders

db.orders.aggregate( [
   {
     $group: {
        _id: null,
        total: { $sum: "$price" }
     }
   }
] )
Sum the price field from orders
SELECT cust_id,
       SUM(price) AS total
FROM orders
GROUP BY cust_id

db.orders.aggregate( [
   {
     $group: {
        _id: "$cust_id",
        total: { $sum: "$price" }
     }
   }
] )
For each unique cust_id, sum the price field.
SELECT cust_id,
       SUM(price) AS total
FROM orders
GROUP BY cust_id
ORDER BY total

db.orders.aggregate( [
   {
     $group: {
        _id: "$cust_id",
        total: { $sum: "$price" }
     }
   },
   { $sort: { total: 1 } }
] )
For each unique cust_id, sum the price field, results sorted by sum.
SELECT cust_id,
       ord_date,
       SUM(price) AS total
FROM orders
GROUP BY cust_id,
         ord_date

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" }
     }
   }
] )
For each uniquecust_id, ord_dategrouping,  sum the pricefield.  Excludes the time portion of the date.
SELECT cust_id,
       count(*)
FROM orders
GROUP BY cust_id
HAVING count(*) > 1

db.orders.aggregate( [
   {
     $group: {
        _id: "$cust_id",
        count: { $sum: 1 }
     }
   },
   { $match: { count: { $gt: 1 } } }
] )
For cust_id with multiple records, return the cust_id and the corresponding record count.
SELECT cust_id,
       ord_date,
       SUM(price) AS total
FROM orders
GROUP BY cust_id,
         ord_date
HAVING total > 250

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" }
     }
   },
   { $match: { total: { $gt: 250 } } }
] )
For each unique cust_id, ord_dategrouping, sum the pricefield  and return only where the sum is greater than 250. Excludes the time portion of the date.
SELECT cust_id,
       SUM(price) as total
FROM orders
WHERE status = 'A'
GROUP BY cust_id

db.orders.aggregate( [
   { $match: { status: 'A' } },
   {
     $group: {
        _id: "$cust_id",
        total: { $sum: "$price" }
     }
   }
] )
For each unique cust_idwith status A, sum the price field.
SELECT cust_id,
       SUM(price) as total
FROM orders
WHERE status = 'A'
GROUP BY cust_id
HAVING total > 250

db.orders.aggregate( [
   { $match: { status: 'A' } },
   {
     $group: {
        _id: "$cust_id",
        total: { $sum: "$price" }
     }
   },
   { $match: { total: { $gt: 250 } } }
] )
For each unique cust_idwith status A, sum the price field and return only where the sum is greater than 250.
SELECT cust_id,
       SUM(li.qty) as qty
FROM orders o,
     order_lineitem li
WHERE li.order_id = o.id
GROUP BY cust_id

db.orders.aggregate( [
   { $unwind: "$items" },
   {
     $group: {
        _id: "$cust_id",
        qty: { $sum: "$items.qty" }
     }
   }
] )
For each unique cust_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: {
               month: { $month: "$ord_date" },
               day: { $dayOfMonth: "$ord_date" },
               year: { $year: "$ord_date"}
           }
        }
     }
   },
   {
     $group: {
        _id: null,
        count: { $sum: 1 }
     }
   }
] )

출처: https://docs.mongodb.com/manual/reference/sql-aggregation-comparison/

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