3

我得到了下表：

文章

+----+-------------+-----------------------------+--------------+
| ID | ID_group_AG | Title                       | Date_publish |
+----+-------------+-----------------------------+--------------+
|  1 |          10 | O obrotach sfer niebieskich | 2009-05-07   |
|  2 |          11 | Technologia betonu          | 2011-03-21   |
|  3 |          12 | test                        | 2008-01-13   |
+----+-------------+-----------------------------+--------------+

员工

+----+-----------+-----------+
| ID | Name      | Surname   |
+----+-----------+-----------+
|  1 | Andrzej   | Gacek     |
|  2 | Leszek    | Ksiazek   |
|  3 | Krzysztof | Skibinski |
|  4 | Andrzej   | Inny      |
+----+-----------+-----------+

文章组

+----+----------+---------------+----------------+
| ID | ID_group | ID_employee   | Points         |
+----+----------+---------------+----------------+
|  1 |       10 |             1 |              3 |
|  2 |       10 |             3 |              3 |
|  3 |       11 |             1 |              2 |
|  4 |       11 |             2 |              2 |
|  5 |       11 |             4 |              2 |
|  6 |       12 |             4 |              6 |
+----+----------+---------------+----------------+

以及以下关系：

articles.ID_group_AG => articlesGroup.ID_group

articlesGroup.ID_employee => employee.ID

我需要做的是打印与员工、文章和发布日期相关的所有文章点，所以我使用以下查询：

SELECT
  p.Name,
  p.Surname,
  a.Date_publish,
  ag.Points
FROM
  employee p,
  articles a,
  articlesGroup ag
WHERE
  (ag.ID_group = a.ID_group_AG) AND
  (ag.ID_employee = p.ID)

我得到：

+-----------+-----------+--------------+----------------+
| Name      | Surname   | Date_publish | Points         |
+-----------+-----------+--------------+----------------+
| Andrzej   | Gacek     | 2009-05-07   |              3 |
| Andrzej   | Gacek     | 2011-03-21   |              2 |
| Leszek    | Ksiazek   | 2011-03-21   |              2 |
| Krzysztof | Skibinski | 2009-05-07   |              3 |
| Andrzej   | Inny      | 2011-03-21   |              2 |
| Andrzej   | Inny      | 2008-01-13   |              6 |
+-----------+-----------+--------------+----------------+

现在让我们解决问题:)

我使用 pChart 库制作图表。

我想把关于每个员工的所有日期放在 Y 轴上，在 X 轴上。

因此，员工“Andrzej Gacek”的积分数组将为：[3,2]

对于员工“Krzysztof Skibinski”将是：[3]

和日期数组（已排序）：["2008-01-13","2009-05-07","2011-03-21"]

我需要在员工点数组中添加零点，例如。对于“Andrzej Gacek”数组应如下所示：[0,3,2]，因此 Point 将与 Dates 相关。

如何形成查询以将零添加到点，以便查询的输出如下所示：

+-----------+-----------+--------------+----------------+
| Name      | Surname   | Date_publish | Points         |
+-----------+-----------+--------------+----------------+
| Andrzej   | Gacek     | 2009-05-07   |              3 |
| Andrzej   | Gacek     | 2011-03-21   |              2 |
| Andrzej   | Gacek     | 2008-01-13   |              0 |
| Leszek    | Ksiazek   | 2011-03-21   |              2 |
| Leszek    | Ksiazek   | 2009-05-07   |              0 |
| Leszek    | Ksiazek   | 2008-01-13   |              0 |
| Krzysztof | Skibinski | 2009-05-07   |              3 |
| Krzysztof | Skibinski | 2011-03-21   |              0 |
| Krzysztof | Skibinski | 2008-01-13   |              0 |
| Andrzej   | Inny      | 2011-03-21   |              2 |
| Andrzej   | Inny      | 2008-01-13   |              6 |
| Andrzej   | Inny      | 2009-05-07   |              0 |
+-----------+-----------+--------------+----------------+
4

1 回答 1

3

您必须创建笛卡尔积才能获得所需的结果。我用 aCROSS JOIN来获取每个员工的日期。

试试这个：

SELECT DISTINCT E.Name,
  E.SurName,
  a.Date_Publish, 
  IFNULL(AG.Points,0) Points
FROM Articles A CROSS JOIN
  Employee E LEFT JOIN
  ArticlesGroup AG ON ag.ID_group = a.ID_group_AG 
    AND E.Id = ag.ID_employee

这是SQL Fiddle。

于 2013-02-18T20:21:29.573 回答