Oracle SQL辅佐计较两种差异的方法,但具有配合的分组

我有一个表,个中包括每个条目标incident_id,status(好比打开或封锁,date_raised(Date)和closure_date(Date).

我想表现一个表来计较封锁日期封锁的变乱数(因此,个中status =’closed’且closure_date不为null的incident_id的计数),以及保持打开的变乱数(events_id的状态,个中状态=统一天’打开’.

万一我狐疑了你,一个看起来像这样的表：

______________________________________________________________________________
| closure date | count of incidents closed | count of incidents remaining open |
|--------------|---------------------------|-----------------------------------|
| 01-Sep-12    | 5                         | 14                                |
| ...          | ...                       | ...                               |

我已经打点了一个表,它将封锁的变乱计数如下：

SELECT COUNT(incident_id)
WHERE closure_date IS NOT NULL AND status="open"
GROUP BY closure_date

我已经实行了几个小时来让其他计数事变,但不能到今朝为止:-(

编辑：这是我有一个表的例子：

___________________________________________________
| incident_id | status | date_raised | closure_date |
|-------------|--------|-------------|--------------|
| 1           | closed | 01-Sep-12   | 01-Sep-12    |
| 2           | open   | 30-Aug-12   | (null)       |
| 3           | open   | 02-Sep-12   | (null)       |
| 4           | closed | 02-Sep-12   | 05-Sep-12    |
| ...         | ...    | ...         | ...          |

会给表：

______________________________________________________________________________
| closure date | count of incidents closed | count of incidents remaining open |
|--------------|---------------------------|-----------------------------------|
| 01-Sep-12    | 1                         | 1                                 |
| 05-Sep-12    | 1                         | 2                                 |

办理要领

SELECT t1.closure_date,COUNT(t1.incident_id),( SELECT COUNT(t2.incident_id) FROM incident_table t2
          WHERE t2.status = 'open'
            AND t2.raised_date < t1.closure_date )
  FROM incident_table t1
 WHERE t1.closure_date IS NOT NULL
   AND t1.status = 'closed'
 GROUP BY t1.closure_date

（编辑：湖南网）

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