按日期分组事件

Question

我有这个用于事件的Oracle表

CREATE TABLE EVENT(
  ID INTEGER NOT NULL,
  SOURCE VARCHAR2(50 ),
  TYPE VARCHAR2(50 ),
  EVENT_DATE DATE,
  DESCRIPTION VARCHAR2(100 )
)
/

我使用此SQL查询按类型获取数据：

SELECT EVENT_DATE, SUM(CASE WHEN TYPE = 'Error' THEN 1 ELSE 0 END) AS ERROR, 
SUM(CASE WHEN TYPE = 'Warning' THEN 1 ELSE 0 END) AS WARN, 
SUM(CASE WHEN TYPE = 'Info' THEN 1 ELSE 0 END) AS INFO, 
SUM(CASE WHEN TYPE = 'Critical' THEN 1 ELSE 0 END) AS CRITICAL
FROM EVENT E WHERE EVENT_DATE >= TRUNC(SYSDATE) - 7 
GROUP BY EVENT_DATE 
HAVING COUNT(CASE WHEN TYPE IN ('Info', 'Warning', 'Error', 'Critical') THEN 1 END) > 0
ORDER BY EVENT_DATE

我如何才能获得一天的结果？我想让所有活动连续7天分开。我应该有7行数据结果

Answer 1

当然David Bachmann Jeppesen的回答是正确的，但您的查询可以简化。

HAVING条件可以在WHERE中应用，在聚合之前减少行数：

SELECT trunc(EVENT_DATE) event_date,
   SUM(CASE WHEN TYPE = 'Error' THEN 1 ELSE 0 END) AS ERROR,
   SUM(CASE WHEN TYPE = 'Warning' THEN 1 ELSE 0 END) AS WARN,
   SUM(CASE WHEN TYPE = 'Info' THEN 1 ELSE 0 END) AS INFO,
   SUM(CASE WHEN TYPE = 'Critical' THEN 1 ELSE 0 END) AS CRITICAL
FROM EVENT E
WHERE EVENT_DATE >= TRUNC(SYSDATE) - 7
  AND TYPE IN ('Info', 'Warning', 'Error', 'Critical')
GROUP BY trunc(EVENT_DATE)
ORDER BY trunc(EVENT_DATE)

Answer 2

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