我想计算Oracle表中的平均值
CREATE TABLE AGENT_HISTORY(
EVENT_ID INTEGER NOT NULL,
AGENTID INTEGER NOT NULL,
EVENT_DATE DATE NOT NULL
)
/
CREATE TABLE CPU_HISTORY(
CPU_HISTORY_ID INTEGER NOT NULL,
EVENT_ID INTEGER NOT NULL,
CPU_NAME VARCHAR2(50 ) NOT NULL,
CPU_VALUE NUMBER NOT NULL
)
/
我使用这个SQL查询:
----- FOR 24 HOURS CPU
CURSOR LAST_24_CPU_CURSOR IS
--SELECT EVENT_DATE, CPU FROM AGENT_HISTORY WHERE NAME = NAMEIN AND EVENT_DATE >= SYSDATE-(60*24)/1440;
SELECT START_DATE, NVL(AVG(CH.CPU_VALUE),0)
FROM (SELECT START_DATE - (LVL+1)/24 START_DATE, START_DATE - LVL/24 END_DATE
FROM (SELECT SYSDATE START_DATE, LEVEL LVL FROM DUAL CONNECT BY LEVEL <= 24))
LEFT JOIN AGENT_HISTORY AH ON EVENT_DATE BETWEEN START_DATE AND END_DATE
LEFT JOIN CPU_HISTORY CH ON AH.EVENT_ID = CH.EVENT_ID
JOIN AGENT AG ON AH.AGENTID = AG.ID
WHERE AG.NAME = NAMEIN
GROUP BY START_DATE
ORDER BY 1;
此查询仅打印一个平均值。我想修改它以打印每小时平均值24个值。你能帮我修改一下这个问题吗?
答案 0 :(得分:1)
我猜你的输入只包含给定间隔之一的数据;由于您正在使用带有AGENT的INNER JOIN
,而AGENT又过滤了AGENT_HISTORY,因此您可以有效地将所有LEFT JOIN
转换为内部版本。
我建议你在AGENT和时间段之间使用CROSS JOIN
代替:
with agent_history(event_date, agentid, event_id) as (
select timestamp '2015-11-18 09:00:07', 1, 1001 from dual
),
agent(id, name) as (
select 1, 'myAgent' from dual
),
cpu_history(event_id, cpu_value) as (
select 1001, 75.2 from dual
),
time_slots(start_date, end_date) as (
SELECT START_DATE - (LVL + 1) / 24 START_DATE,
START_DATE - LVL / 24 END_DATE
FROM (SELECT SYSDATE START_DATE,
LEVEL LVL
FROM DUAL
CONNECT BY LEVEL <= 24)
)
SELECT START_DATE,
NVL(AVG(CH.CPU_VALUE),
0)
FROM time_slots ts
CROSS JOIN AGENT AG
LEFT JOIN AGENT_HISTORY AH
ON AH.AGENTID = AG.ID
AND EVENT_DATE BETWEEN START_DATE AND END_DATE
LEFT JOIN CPU_HISTORY CH
ON AH.EVENT_ID = CH.EVENT_ID
WHERE AG.NAME = 'myAgent'
GROUP BY START_DATE
ORDER BY 1;
这可确保您获得完整的24行(每个时间段一行)。
答案 1 :(得分:0)
在start_date
和to_char(start_date, 'hh24:mi')
条款中将select
更改为group by
。