SQL subquery to get the total

问题

Using SQL subquery, how do I get the total items and total revenue for each manager including his team? Suppose I have this table items_revenue with columns:

All the managers (is_manager=1) and their respective members are in the above table. Member1 is under Manager1, Member2 is under Manager2, and so on, but real data are in random arrangement.

I want my query to output the ff.:

This is related to SQL query to get the subtotal of some rows but I don't want to use the CASE expression. Thanks!

You can copy this to easily create the table:

DROP TABLE IF EXISTS items_revenue;
CREATE TABLE items_revenue (id int, is_manager int, manager_id int, name varchar(55), no_of_items int, revenue int);
INSERT INTO items_revenue (id, is_manager, manager_id, name, no_of_items, revenue)
VALUES
    (1  ,    1     ,    0     , 'Manager1' ,    621    ,   833), 
    (2  ,    1     ,    0     , 'Manager2' ,    458    ,   627),
    (3  ,    1     ,    0     , 'Manager3' ,    872    ,   1027 ),
    (8  ,    0     ,    1     , 'Member1'  ,    1258   ,   1582),
    (9  ,    0     ,    2     , 'Member2'  ,    5340   ,   8827),
    (10  ,    0     ,    3     , 'Member3'  ,    3259   ,   5124);

回答1:

Use union all and aggreation:

select manager_id, sum(no_of_items) as no_of_items, sum(revenue) as revenue
from ((select ir.manager_id, ir.no_of_items, ir.revenue
       from items_revenue ir
       where ir.is_manager = 0
      ) union all
      (select ir.id, ir.no_of_items, ir.revenue
       from items_revenue ir
       where ir.is_manager = 1
      )
     ) irm
group by manager_id;

Note: This only handles direct reports in the table. This is the sample data that you provide. The problem is significantly different if you need all direct reports, so don't modify this question for that situation (ask another). If that is your need, then MySQL is not the best tool (unless you are using version 8), although you can solve it if you know the maximum depth.

回答2:

Use two subqueries, one grouped by id where is_manager=1 and another grouped by manager_id where is_manager=0.

SELECT id, name, 
       (t1.total_items + t2.total_items) total_items,
       (t1.total_revenue + t2.total_revenue) total_revenue
FROM (SELECT id, name, 
             sum(no_of_items) total_items, 
             sum(revenue) total_revenue
      FROM   items_revenue
      WHERE  is_manager = 1
      GROUP BY id, name) t1
JOIN (SELECT manager_id,
             sum(no_of_items) total_items, 
             sum(revenue) total_revenue
      FROM   items_revenue
      WHERE  is_manager = 0
      GROUP BY manager_id) t2
ON    t1.id = t2.manager_id;

Rextester here

回答3:

SELECT
  employee.totalit + items_revenue.no_of_items,
  employee.totalerev + items_revenue.revenue,
  employee.manager_id
FROM (SELECT
        sum(no_of_items) AS totalit,
        sum(revenue) as totalerev,
        manager_id
      FROM items_revenue
      WHERE manager_id <> 0
      GROUP BY manager_id) AS employee, items_revenue
WHERE items_revenue.user_id = employee.manager_id;

来源：https://stackoverflow.com/questions/49383338/sql-subquery-to-get-the-total

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