select count DISTINCT where！= 2 filter

Question

我的表格有 eventid （警报事件ID）和事件类型（active = 1, Ack = 2, unack = 0）。我想展示一些未经确认的警报。

SELECT count( DISTINCT eventid) FROM alarm_events where eventtype != 2

不确定我如何过滤掉已被确认的黄昏

Answer 1

SELECT COUNT(DISTINCT eventid)
FROM alarm_events
WHERE eventid NOT IN (
    SELECT eventid
    FROM alarm_events
    WHERE eventtype = 2)

或等效

SELECT COUNT(DISTINCT a.eventid)
FROM alarm_events AS a
LEFT JOIN alarm_events AS b ON a.eventid = b.eventid AND b.eventtype = 2
WHERE b.eventid IS NULL

请参阅Return row only if value doesn't exist

Answer 2

SELECT count( DISTINCT eventid) FROM alarm_events where eventtype in (1,0)

试试这个

Answer 3

您可以使用分组依据进行操作，对于过滤，您可以使用IN运算符。

SELECT count(eventid) FROM alarm_events WHERE eventtype in (0, 1) GROUP BY eventid

  

如果您的eventtype列是String，则使用（'0'，'1'）替换（0,1）。