Question

我有一个包含以下列的表：

|customerNum|OrderNumber|ArticleNumber|PaymentType|Returned?|OrderDate |
|1          |1          |1            |cash       |Yes      |01.01.2016|
|2          |2          |3            |credit     |No       |30.02.2016|
|1          |1          |2            |cash       |No       |01.01.2016|
|2          |2          |4            |credit     |Yes      |30.02.2016|
|3          |3          |2            |credit     |No       |17.05.2016|
|2          |4          |8            |cash       |Yes      |19.08.2016|

预期结果：

customerNum|Numb.ofOrdersTot|ratioCash/CreditPayment|returnRate|LastOrder |
|1         |1               |1                      |0,5       |01.01.2016|
|2         |2               |0.5                    |0,66      |19.08.2016|
|3         |1               |0                      |0         |17.05.2016|

我想计算每个客户的订单总数（非订购商品）（一个订单可以有多个商品），现金支付和信用支付的比率（现金支付数量/信用支付数量（一个）订单与多个商品具有相同的付款方式））客户的退货率（numberOfReturnedArticles / numberOfArticlesOrderedbycustomerTotal）和最后订单日期

我已经尝试了它的一部分，但这不起作用而且不完整：

`COUNT(*) AS Numb.ofOrdersTot,(COUNT(orderNumber)/COUNT(Returned? = Yes)) FROM table 1 group by customerNum;`

PLZ帮助我：非常重要：/

Answer 1

这是我的答案。请检查它，看看字段是否正确类型。此外，我已添加2秒字段'orderdate2'与另一种格式。你可以删除一个。所以你的'预期结果'不能来自你的输入

SELECT
    customerNum AS 'customerNum'
    , SUM(1) AS 'Numb.ofOrdersTot'
    , SUM(IF(`PaymentType` = 'credit',1,0)) / SUM(1) AS 'ratioCash/CreditPayment'
    , SUM(IF(`Returned?` = 'Yes',1,0)) / SUM(1) AS 'returnRate'
    ,MAX(`OrderDate`) AS 'LastOrder'
    ,DATE_FORMAT(MAX(`OrderDate`),'%d.%m.%Y') AS 'LastOrder2'   
FROM orders
GROUP BY customerNum;

<强>样品

mysql> SELECT * FROM orders;
+----+-------------+-------------+---------------+-------------+-----------+------------+
| id | customerNum | OrderNumber | ArticleNumber | PaymentType | Returned? | OrderDate  |
+----+-------------+-------------+---------------+-------------+-----------+------------+
|  1 |           1 |           1 |             1 | cash        | Yes       | 2016-01-01 |
|  2 |           2 |           2 |             3 | credit      | No        | 2016-02-29 |
|  3 |           1 |           1 |             2 | cash        | No        | 2016-01-01 |
|  4 |           2 |           2 |             4 | credit      | Yes       | 2016-02-29 |
|  5 |           3 |           3 |             2 | credit      | No        | 2016-05-17 |
|  6 |           2 |           4 |             8 | cash        | Yes       | 2016-08-19 |
+----+-------------+-------------+---------------+-------------+-----------+------------+
6 rows in set (0,00 sec)

mysql> SELECT
    ->     customerNum AS 'customerNum'
    ->     , SUM(1) AS 'Numb.ofOrdersTot'
    ->     , SUM(IF(`PaymentType` = 'credit',1,0)) / SUM(1) AS 'ratioCash/CreditPayment'
    ->     , SUM(IF(`Returned?` = 'Yes',1,0)) / SUM(1) AS 'returnRate'
    ->     ,MAX(`OrderDate`) AS 'LastOrder'
    ->     ,DATE_FORMAT(MAX(`OrderDate`),'%d.%m.%Y') AS 'LastOrder2'
    -> FROM orders
    -> GROUP BY customerNum;
+-------------+------------------+-------------------------+------------+------------+------------+
| customerNum | Numb.ofOrdersTot | ratioCash/CreditPayment | returnRate | LastOrder  | LastOrder2 |
+-------------+------------------+-------------------------+------------+------------+------------+
|           1 |                2 |                  0.0000 |     0.5000 | 2016-01-01 | 01.01.2016 |
|           2 |                3 |                  0.6667 |     0.6667 | 2016-08-19 | 19.08.2016 |
|           3 |                1 |                  1.0000 |     0.0000 | 2016-05-17 | 17.05.2016 |
+-------------+------------------+-------------------------+------------+------------+------------+
3 rows in set (0,00 sec)

mysql>

使用MySQL计算每个客户的订单数量，paymentRatio，退货率和最后订单日期

1 个答案: