使用MySQL生成每日销售报告，填补空白，按货币分组

时间：2010-04-29 20:36:11

标签： mysql join report mysql5

我正在尝试使用MySQL 5.1.45创建我认为是在线商店的相对基本的报告

商店可以接收多种货币付款。 我已经创建了一些带有数据的示例表，并且我试图生成一个按日期和货币分组的简单表格结果集，以便我可以绘制这些数字。

我想查看每个日期可用的每种货币，如果当天没有该货币的销售，则结果为0。 如果我可以开始工作，我想做同样的事情，但也按照产品ID进行分组。

在我提供的示例数据中，只有3种货币和2种产品ID，但实际上每种货币可以有任意数量。

我可以按日期正确分组，但是当我按货币添加分组时，我的查询不会返回我想要的内容。

我的工作基于this article。

我的报告查询，仅按日期分组：

SELECT calendar.datefield AS date,
   IFNULL(SUM(orders.order_value),0) AS total_value
FROM orders 
RIGHT JOIN calendar ON (DATE(orders.order_date) = calendar.datefield)
WHERE (calendar.datefield BETWEEN (SELECT MIN(DATE(order_date)) FROM orders) AND  (SELECT MAX(DATE(order_date)) FROM orders))
GROUP BY date

现在按日期和货币分组：

SELECT calendar.datefield AS date, orders.currency_id,
   IFNULL(SUM(orders.order_value),0) AS total_value
FROM orders 
RIGHT JOIN calendar ON (DATE(orders.order_date) = calendar.datefield)
WHERE (calendar.datefield BETWEEN (SELECT MIN(DATE(order_date)) FROM orders) AND (SELECT MAX(DATE(order_date)) FROM orders))
GROUP BY date, orders.currency_id

我得到的结果（按日期和货币分组）：

+------------+-------------+-------------+
| date       | currency_id | total_value |
+------------+-------------+-------------+
| 2009-08-15 |           3 |       81.94 |
| 2009-08-15 |          45 |       25.00 |
| 2009-08-15 |          49 |      122.60 |
| 2009-08-16 |        NULL |        0.00 |
| 2009-08-17 |          45 |       25.00 |
| 2009-08-17 |          49 |      122.60 |
| 2009-08-18 |           3 |       81.94 |
| 2009-08-18 |          49 |      245.20 |
+------------+-------------+-------------+

我想要的结果：

+------------+-------------+-------------+
| date       | currency_id | total_value |
+------------+-------------+-------------+
| 2009-08-15 |           3 |       81.94 |
| 2009-08-15 |          45 |       25.00 |
| 2009-08-15 |          49 |      122.60 |
| 2009-08-16 |           3 |        0.00 |
| 2009-08-16 |          45 |        0.00 |
| 2009-08-16 |          49 |        0.00 |
| 2009-08-17 |           3 |        0.00 |
| 2009-08-17 |          45 |       25.00 |
| 2009-08-17 |          49 |      122.60 |
| 2009-08-18 |           3 |       81.94 |
| 2009-08-18 |          45 |        0.00 |
| 2009-08-18 |          49 |      245.20 |
+------------+-------------+-------------+

我在测试中使用的架构和数据：

CREATE TABLE orders
(
  id INT PRIMARY KEY AUTO_INCREMENT,
  order_date DATETIME,
  order_id INT,
  product_id INT,
  currency_id INT,
  order_value DECIMAL(9,2),
  customer_id INT
);
INSERT INTO orders (order_date, order_id, product_id, currency_id, order_value, customer_id)
  VALUES
  ('2009-08-15 10:20:20', '123', '1', '45', '12.50', '322'),
  ('2009-08-15 12:30:20', '124', '1', '49', '122.60', '400'),
  ('2009-08-15 13:41:20', '125', '1', '3', '40.97', '324'),  
  ('2009-08-15 10:20:20', '126', '2', '45', '12.50', '345'),
  ('2009-08-15 13:41:20', '131', '2', '3', '40.97', '756'),

  ('2009-08-17 10:20:20', '3234', '1', '45', '12.50', '1322'),
  ('2009-08-17 10:20:20', '4642', '2', '45', '12.50', '1345'),
  ('2009-08-17 12:30:20', '23', '2', '49', '122.60', '3142'),

  ('2009-08-18 12:30:20', '2131', '1', '49', '122.60', '4700'),
  ('2009-08-18 13:41:20', '4568', '1', '3', '40.97', '3274'),  
  ('2009-08-18 12:30:20', '956', '2', '49', '122.60', '3542'),
  ('2009-08-18 13:41:20', '443', '2', '3', '40.97', '7556');

CREATE TABLE currency
  (
    id INT PRIMARY KEY,
    name VARCHAR(255)
  );
INSERT INTO currency (id, name)
  VALUES
  (3, 'Euro'),
  (45, 'US Dollar'),
  (49, 'CA Dollar');


CREATE TABLE calendar (datefield DATE);

  DELIMITER |
  CREATE PROCEDURE fill_calendar(start_date DATE, end_date DATE)
  BEGIN
    DECLARE crt_date DATE;
    SET crt_date=start_date;
    WHILE crt_date < end_date DO
      INSERT INTO calendar VALUES(crt_date);
      SET crt_date = ADDDATE(crt_date, INTERVAL 1 DAY);
    END WHILE;
  END |
  DELIMITER ;

CALL fill_calendar('2008-01-01', '2011-12-31');

2 个答案:

答案 0 :(得分：3)

你会发现很难得到你想要的结果，除非你在每天为每种货币输入一个虚拟订单（这可以在fill_calendar程序中轻松完成以进行测试）。

目前，您想要的是使用公共链接加入日历，订单和货币;但是没有这样的链接（你有从日历到订单和订单到货币的链接，但从日历到货币都没有）。

如果你创建了这些虚拟订单，那么你就不需要改变架构了;数据本身将提供所需的链接。否则，您可能需要稍微更改架构。

答案 1 :(得分：2)

我发布这个作为答案，因为评论可能会非常大。 感谢Mark指出我正确的方向。 Mark的答案有效，但意味着对日历表的架构更改，我并不热衷于此，因为我可能需要将来更灵活的报告（例如， product_id ）

这有效 - 但它可能不优雅。我会将这个问题留下“未答复”几天，以确定是否有人能提出更好的解决方案。

其他架构和数据（添加产品表）：

CREATE TABLE products
  (
    id INT PRIMARY KEY,
    name VARCHAR(255)
  );
INSERT INTO products (id, name)
  VALUES
  (1, 'Widget'),
  (2, 'Midget'),
  (3, 'Gidget');

现在使用此查询我得到了我想要的答案：

SELECT cal.date AS date, currency.name AS currency, products.name AS product,
   IFNULL(SUM(orders.order_value),0) AS total_value
FROM orders 
RIGHT JOIN 
(
SELECT cal.datefield AS date, cur.id AS currency, prod.id AS product
FROM calendar cal
CROSS JOIN currency cur
CROSS JOIN products prod
) cal
 ON (DATE(orders.order_date) = cal.date)
    AND orders.currency_id = cal.currency
    AND orders.product_id = cal.product
JOIN currency ON cal.currency = currency.id
JOIN products ON cal.product = products.id
WHERE (cal.date BETWEEN (SELECT MIN(DATE(order_date)) FROM orders) AND (SELECT MAX(DATE(order_date)) FROM orders))
GROUP BY date, cal.currency,cal.product

这为我提供了所有日期的所有数据点，如果它们不存在则为零。

+------------+-----------+--------+-------------+
| date       | currency  | product| total_value |
+------------+-----------+--------+-------------+
| 2009-08-15 | Euro      | Widget |       40.97 |
| 2009-08-15 | Euro      | Midget |       40.97 |
| 2009-08-15 | Euro      | Gidget |        0.00 |
| 2009-08-15 | US Dollar | Widget |       12.50 |
| 2009-08-15 | US Dollar | Midget |       12.50 |
| 2009-08-15 | US Dollar | Gidget |        0.00 |
| 2009-08-15 | CA Dollar | Widget |      122.60 |
| 2009-08-15 | CA Dollar | Midget |        0.00 |
| 2009-08-15 | CA Dollar | Gidget |        0.00 |
| 2009-08-16 | Euro      | Widget |        0.00 |
| 2009-08-16 | Euro      | Midget |        0.00 |
| 2009-08-16 | Euro      | Gidget |        0.00 |
| 2009-08-16 | US Dollar | Widget |        0.00 |
| 2009-08-16 | US Dollar | Midget |        0.00 |
| 2009-08-16 | US Dollar | Gidget |        0.00 |
| 2009-08-16 | CA Dollar | Widget |        0.00 |
| 2009-08-16 | CA Dollar | Midget |        0.00 |
| 2009-08-16 | CA Dollar | Gidget |        0.00 |
| 2009-08-17 | Euro      | Widget |        0.00 |
| 2009-08-17 | Euro      | Midget |        0.00 |
| 2009-08-17 | Euro      | Gidget |        0.00 |
| 2009-08-17 | US Dollar | Widget |       12.50 |
| 2009-08-17 | US Dollar | Midget |       12.50 |
| 2009-08-17 | US Dollar | Gidget |        0.00 |
| 2009-08-17 | CA Dollar | Widget |        0.00 |
| 2009-08-17 | CA Dollar | Midget |      122.60 |
| 2009-08-17 | CA Dollar | Gidget |        0.00 |
| 2009-08-18 | Euro      | Widget |       40.97 |
| 2009-08-18 | Euro      | Midget |       40.97 |
| 2009-08-18 | Euro      | Gidget |        0.00 |
| 2009-08-18 | US Dollar | Widget |        0.00 |
| 2009-08-18 | US Dollar | Midget |        0.00 |
| 2009-08-18 | US Dollar | Gidget |        0.00 |
| 2009-08-18 | CA Dollar | Widget |      122.60 |
| 2009-08-18 | CA Dollar | Midget |      122.60 |
| 2009-08-18 | CA Dollar | Gidget |        0.00 |
+------------+-----------+--------+-------------+

这在子查询上使用了一个JOIN，我觉得它不是很高效，但它适用于这个小数据集 - 我会生成更多的数据，看看它是怎么回事。

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