Question

我正在构建一个查询以返回每日销售数据。我当前的查询返回一个类似于这样的表：

----------------------------------
| DATE       | SKU      | TOTAL  | 
----------------------------------
| 2014-11-01 | AV155_A  | 209.00 |
| 2014-11-02 | AV155_B  | 627.00 |
| 2014-11-04 | AV155_C  | 279.00 |
| 2014-11-05 | AV155    | 279.00 |
| 2014-11-08 | AV1556_A | 209.00 |
| 2014-11-09 | AV1556_B | 627.00 |
| 2014-11-10 | AV1556_C | 279.00 |
| 2014-11-12 | AV1556   | 279.00 |

我想要的是每天显示的结果表，即使该特定日期没有数据点也是如此。像这样：

----------------------------------
| DATE       | SKU      | TOTAL  | 
----------------------------------
| 2014-11-01 | AV155_A  | 209.00 |
| 2014-11-02 | AV155_B  | 627.00 |
| 2014-11-03 |          |      0 |
| 2014-11-04 | AV155_C  | 279.00 |
| 2014-11-05 | AV155    | 279.00 |
| 2014-11-06 |          |      0 |
| 2014-11-07 |          |      0 |
| 2014-11-08 | AV1556_A | 209.00 |
| 2014-11-09 | AV1556_B | 627.00 |
| 2014-11-10 | AV1556_C | 279.00 |
| 2014-11-11 |          |      0 |
| 2014-11-12 | AV1556   | 279.00 |

我目前的查询如下：

select
  DATE_FORMAT(created_on, '%m-%d-%Y') as date,
  sku,
  SUM(price) as total 
FROM order_items
WHERE created_on between FROM_UNIXTIME(1415577600) AND NOW()
GROUP BY MONTH(created_on), DAY(v.created_on), order_item_sku;

Answer 1

您需要使用外部联接。最简单的方法是如果你有一个日历表，但你可以动态制作一个：

select c.thedate, oi.sku, sum(price) as total
from (select date('2014-11-01') as thedate union all
             date('2014-11-02') as thedate union all
             date('2014-11-03') as thedate union all
             date('2014-11-04') as thedate union all
             date('2014-11-05') as thedate union all
             date('2014-11-06') as thedate union all
             date('2014-11-07') as thedate union all
             date('2014-11-08') as thedate union all
             date('2014-11-09') as thedate union all
             date('2014-11-10') as thedate union all
             date('2014-11-11') as thedate union all
             date('2014-11-12') as thedate
      ) c left join
      order_items oi
      on c.thedate = date(oi.created_on)
where oi.created_on between FROM_UNIXTIME(1415577600) AND NOW()
group by ci.thedate, oi.sku

Answer 2

这是一个解决灵活的日期列表需求的答案。您需要找到一种方法来获取包含适当范围内所有日期的虚拟表，然后将它们连接到摘要。这是一个查询，它将获取范围内的日期。

SELECT mintime + INTERVAL seq.seq DAY AS reportdate
  FROM (
        SELECT MIN(DATE(created_on)) AS mintime,
               MAX(DATE(created_on)) AS maxtime
          FROM order_items
         WHERE created_on >= starting_time
           AND created_on <= NOW()
       ) AS order_items
  JOIN seq_0_to_999 AS seq 
                    ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)

这里发生了什么？三件事。

我们有一个子查询，用于确定我们关心报告的第一天和最后一天（最小和最大created_on）。
我们将时间范围应用于该查询。我希望避免使用BETWEEN作为时间戳范围，因为它经常会在一秒钟的错误中错误地结束时间。
我们有一个名为seq_0_to_999的表。它包含一千个基数的序列：从零开始的整数。稍等一下。

然后，您可以将其作为子查询加入到聚合查询中，以获取列出范围内的所有日期，如此。

select  DATE_FORMAT(d.reportdate, '%m-%d-%Y') as date,
        sku,
        SUM(price) as total 
   FROM (
          SELECT mintime + INTERVAL seq.seq DAY AS reportdate
            FROM (
                  SELECT MIN(DATE(created_on)) AS mintime,
                         MAX(DATE(created_on)) AS maxtime
                    FROM order_items
                   WHERE created_on >= starting_time
                     AND created_on <= NOW()
                 ) AS order_items
            JOIN seq_0_to_999 AS seq 
                              ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)
        ) AS d
   LEFT JOIN order_items ON d.reportdate = DATE(order_items.created_on)       
   WHERE created_on >= starting_time
     AND created_on <= NOW()
   GROUP BY d.reportdate, sku
   ORDER BY d.reportdate, sku

它看起来像一个查询的大讨厌毛球。但如果你把它看成是由各种层次的查询组成的三明治，那真的不是那么复杂。

它使用LEFT JOIN，因此即使您的order_items表中没有相应的数据，也可以确保范围内的所有日期都已保留。

最后，这个seq_0_to_999表怎么样？我们从零开始得到那些整数？答案是：我们必须安排这样做;这些数字不是内置在MySQL中的。（它们内置在名为MariaDB的MySQL分支中。）创建一个包含0-9整数的短表，如下所示：

DROP TABLE IF EXISTS seq_0_to_9;
CREATE TABLE seq_0_to_9 AS
   SELECT 0 AS seq UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
    UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9;

然后创建一个视图，将该表与自身连接起来，生成1000个这样的组合：

DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
  FROM seq_0_to_9 a
  JOIN seq_0_to_9 b
  JOIN seq_0_to_9 c
);

我在http://www.plumislandmedia.net/mysql/filling-missing-data-sequences-cardinal-integers/

MySQL报告 - 填写空日期

2 个答案: