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我正在尝试使用 MySQL 5.1.45 为在线商店创建我认为相对基本的报告

商店可以接收多种货币的付款。我已经创建了一些包含数据的示例表，并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。 如果我能让它工作，我想做同样的事情，但也按产品 ID 分组。

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

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。

我的工作基于这篇文章。

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');
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2 回答 2

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你会发现很难在那里得到你想要的结果，除非你每天在系统中为每种货币输入一个虚拟订单（这可以在 fill_calendar 例程中轻松完成以进行测试）。

现在，您想要的是使用通用链接加入日历、订单和货币；但没有这样的链接（您有从日历到订单和订单到货币的链接，但没有从日历到货币的链接）。

如果您创建了这些虚拟订单，则无需更改架构；数据本身将提供所需的链接。否则，您可能必须稍微更改架构。

于 2010-04-30T11:32:27.103 回答
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我将其发布为答案，因为评论可能很大。感谢 Mark 为我指明了正确的方向。马克的回答有效，但意味着我不热衷于对日历表进行架构更改，因为我将来可能需要报告更加灵活（例如按product_id分组）

这行得通 - 但是它可能并不优雅。我将把这个问题“悬而未决”几天，看看是否有人能提出更好的解决方案。

附加架构和数据（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');

现在使用这个查询，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

这给了我所有日子的所有数据点，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 JOIN，我认为它的性能不是很好，但它适用于这个小数据集 - 我将生成更多数据并看看它是如何进行的。

于 2010-04-30T12:19:39.630 回答