Question

我有三张桌子：

// priorities       // statuses              // projects
+----+--------+     +----+-------------+     +----+------+--------+----------+
| ID | NAME   |     | ID | STATUS NAME |     | ID | NAME | STATUS | PRIORITY |
+----+--------+     +----+-------------+     +----+------+--------+----------+
| 1  | Normal |     | 1  | Pending     |     | 1  | a    | 1      | 3        |
+----+--------+     +----+-------------+     +----+------+--------+----------+
| 2  | High   |     | 2  | In Progress |     | 2  | b    | 1      | 1        |
+----+--------+     +----+-------------+     +----+------+--------+----------+
| 3  | Urgent |                              | 3  | c    | 2      | 1        |
+----+--------+                              +----+------+--------+----------+
                                             | 4  | d    | 1      | 2        |
                                             +----+------+--------+----------+

我需要创建一个view，其中显示有projects个status 1和1的优先级，有多少人status {1}} 1，优先级2，有status 1，优先级3，等等。< / p>

这应该遍历每个状态，然后是每个优先级，然后计算适用于标准的项目。

视图应该包含如下值：

// VIEW (stats)
+--------+----------+-------+
| STATUS | PRIORITY | COUNT |
+--------+----------+-------+
| 1      | 1        | 1     |
+--------+----------+-------+
| 1      | 2        | 1     |
+--------+----------+-------+
| 1      | 3        | 1     |
+--------+----------+-------+
| 2      | 1        | 1     |
+--------+----------+-------+
| 2      | 2        | 0     |
+--------+----------+-------+
| 2      | 3        | 0     |
+--------+----------+-------+

此观点是，我可以调用，例如，有多少项目的状态为1，优先级为3，给出上述数据的答案应为1

使用下面的select语句我已经能够产生类似的结果，但它没有明确显示0项目的状态为2且优先级为{{1} }。我需要使用此3值，与0的任何其他值相同。

COUNT >= 1

我怎样才能创建一个包含所有优先级的VIEW＆＃39; ID，所有状态＆＃39; // my current select statement CREATE VIEW stats AS SELECT P.STATUS, P.PRIORITY, COUNT(*) AS hits FROM projects P GROUP BY P.STATUS, P.PRIORITY // does not show rows where COUNT = 0的ID，和 0值？

Answer 1

您需要生成所有行，然后获取每个行的计数。这是一个应该有效的查询：

SELECT s.status, p.priority, COUNT(pr.status) AS hits
FROM (SELECT DISTINCT status FROM projects) s CROSS JOIN
     (SELECT DISTINCT priority FROM projects) p LEFT JOIN
     project pr
     ON pr.status = s.status and pr.priority = p.priority
GROUP BY s.status, p.priority;

使用多个表创建视图 - 必须显示0个值

1 个答案: