如何计算数量订单的最大值、最小值、平均值和中值

Question

我有一个名为 order_match 的 table，其中包含 order_buyer_Id 作为交易 ID，createdby 作为买家 ID，createdAt 作为交易发生的日期和数量作为每个订单的数量。

在这种情况下，我想计算每个买家（创建者）的订单 (order_buyer_Id)，然后找出最大和最小数量。

这是示例数据：

+----------------+-----------+------------+--------+
| order_buyer_id | createdby | createdAt  |quantity|
+----------------+-----------+------------+--------+
|          19123 |        19 | 2017-02-02 |0.4     |
|         193241 |        19 | 2017-02-02 |0.5       
|         123123 |        20 | 2017-02-02 |1       |
|          32242 |        20 | 2017-02-02 |4
|          32434 |        20 | 2017-02-02 |3       |
+----------------+-----------+------------+---------

如果我运行查询，预期结果是：

+-----+-----+---------+--------+
| max | min | average | median |
+-----+-----+---------+--------+
|   4 | 0.4 |    1,78 |      1 |
+-----+-----+---------+---------

这是fiddle

http://www.sqlfiddle.com/#!9/d89772/15

这是我的查询

SELECT MAX(quantity) AS max,
       MIN(quantity) AS min,
       AVG(quantity) AS average,
       AVG(CASE WHEN rn IN (FLOOR((@tr+1)/2), FLOOR((@tr+2)/2)) THEN quantity END) AS median
FROM (
  SELECT count, 
         @rn := @rn + 1 AS rn,
         @tr := @rn AS tr
  FROM (
    SELECT COUNT(*) AS count
    FROM order_match
    GROUP BY order_buyer_Id
    order by quantity
  ) o
  CROSS JOIN (SELECT @rn := 0) init
) c

Answer 1

SELECT t.max,t.min,t.average,0.00 AS 'Median'

FROM
(SELECT MAX(quantity) AS max,

       MIN(quantity) AS min,

       SUM(quantity)/COUNT(distinct created_by) AS average

FROM order_match)t

union 

SELECT 0.00 AS 'max',0.00 AS 'min',0.00 AS 'Average',

((2*t1.average/3)+t1.mode) AS 'Median' 

FROM (SELECT count(FLOOR(quantity)),IFNULL(FLOOR(quantity),min(quantity)) AS 'mode'  

FROM order_match GROUP BY quantity HAVING 

count(FLOOR(quantity))>1)t1

Answer 2

您收到错误消息是因为 quantity 不在您的 subquery 中。

您可以再次加入您的 table 以获得 quantity 或者您可以在 select 中包含数量（甚至根据您的示例数据group by 与数量给出相同的结果)

SELECT MAX(quantity) AS max,
       MIN(quantity) AS min,
       AVG(quantity) AS average,
       AVG(CASE WHEN rn IN (FLOOR((@tr+1)/2), FLOOR((@tr+2)/2)) THEN quantity END) AS median
FROM (
  SELECT count, quantity,
         @rn := @rn + 1 AS rn,
         @tr := @rn AS tr
  FROM (
    SELECT COUNT(*) AS count,Quantity
    FROM order_match
    GROUP BY order_buyer_Id,Quantity
    order by quantity
  ) o
  CROSS JOIN (SELECT @rn := 0) init
) c

SQL FIDDLE

如何计算数量订单的最大值、最小值、平均值和中值

how to count max, min, average, and median of quantity order

mysql

mysql-workbench