Question

我在Zabbix数据库中有值，每分钟加载到它上面。我需要一个计算不同日期的小时总计的查询，例如2013年3月20日，我有一行0:00，其中包含时间＆gt; = 0:00和＆lt;凌晨1点等。然后我会在上午1点到凌晨2点之间有另一行。我正在使用下面的查询，但我必须不断改变时间。我正在寻找的是一个查询，它将为我生成一天中每小时的24行。请帮忙。

SELECT 
    SUM(CASE WHEN itemid = 23661
        THEN value ELSE 0 END) Hits 
    FROM history_uint WHERE 
        clock >= EXTRACT(EPOCH FROM TIMESTAMP '2013-03-24 00:00:00')
         AND clock < EXTRACT(EPOCH FROM TIMESTAMP '2013-03-24 01:00:00')

Answer 1

您是否尝试按小时分组结果？

SELECT DATEPART(HOUR, timestamp) [HOUR]
, SUM(CASE WHEN itemid = 23661 THEN value ELSE 0 END) hits 
FROM history_uint
WHERE clock >= EXTRACT(EPOCH FROM timestamp '2013-03-24')
  AND clock < EXTRACT(EPOCH FROM timestamp '2013-03-25')
GROUP BY DATEPART(HOUR, timestamp);

Answer 2

类似的东西：

select '2013-03-20'::date + delta * '1 hour'::interval
from generate_series(0,23) g(delta)

Answer 3

select
    date_trunc('hour', clock) "hour",
    sum((itemid = 23661)::integer) hits 
from history_uint
group by 1
order by 1

或者填满所有时间：

select
    s.hour, count(itemid = 23661 or null) hits 
from
    (
        select date_trunc('hour', clock) "hour", itemid
        from history_uint
    ) h
    right join (
        select date_trunc('hour', d) "hour"
        from generate_series (
            (select min(clock::date)),
            (select max(clock)::date + 1) - interval '1 hour',
            '1 hour'
        ) s(d)
    ) s on s.hour = h.hour
group by 1
order by 1

SQL中的迭代求和

3 个答案: