按时间汇总记录

Question

我有一个查询来按时间执行聚合记录，但我有一些错误。

我的情况：我将MySQL db 5.6中的股票市场数据存储在从0900到1730的每分钟一条记录中，即每天511条记录

我需要在不同的时间范围内汇总这些数据，假设为5分钟，那么

volume5min - ＆gt;总量0900：0904

open5min - ＆gt; 0900记录的开放值（范围内的第一个开放值）

close5min - ＆gt; 0904记录的近似值（范围内的最后一个接近值）

high5min - ＆gt; 0900至0904范围内的最高值

low5min - ＆gt; 0900至0904范围内的最低值

等等。

我有一个查询来完成此操作，但是我在打开和关闭聚合值时出现错误

SELECT 
    floor(cast(time as SIGNED) / 5) as timeInterval,
    date,
    time,
    MAX(high) AS high,
    MIN(low) as low,
    SUM(volume) as volume,
    (select 
            open
        from
            atlantia a2
        where
            a1.time = a2.time
        order by time
        limit 1) as open,
    (select 
            close
        from
            atlantia a2
        where
            a1.time = a2.time
        order by time desc
        limit 1) as close
FROM
    atlantia a1
GROUP BY date , timeInterval

这是我执行查询的原因

打开和关闭没有正确聚合，而其他列似乎很好。

更重要的是，更改总时间范围我得到的时间不正确，例如下面的60分钟示例

从0900开始我应该有1000,1100等等，而现在我也有1020,1140等。

列类型为：

日期：CHAR 时间：CHAR 所有其余的都是DOUBLE但是Volume是INTEGER。

如何修改此查询以正确汇总值？

编辑：为了验证您的上一次查询，我手动检查了任何一个小时，这些是我应该得到的正确值

您的查询会返回关闭值的差异，其余的都没关系

关闭值应该是时间范围的最后一条记录，即0900和0959之间的时间关闭是关闭列中的0959值。

编辑2：看来我已经找到了诀窍的所在，现在有了这个查询，一切正常

 SELECT 
 Sub1.timeInterval, 
 a1.date,
 MIN(a1.time),
 MAX(a1.high) AS high,
 MIN(a1.low) as low,
 SUM(a1.volume) as volume,
 a2.open as open,
 a3.close as close
 FROM atlantia a1
 INNER JOIN
 (
 SELECT floor( (cast( SUBSTRING(time,1,2) AS SIGNED ) * 60 + cast( SUBSTRING(time,3,2)   AS SIGNED )) /60 ) AS timeInterval, MIN(time) AS minTime, MAX(time) AS maxtime
 FROM atlantia
 GROUP BY timeInterval
 ) Sub1
 ON floor( (cast( SUBSTRING(a1.time,1,2) AS SIGNED ) * 60 + cast( SUBSTRING(a1.time,3,2)    AS SIGNED )) /60 ) = Sub1.timeInterval
 INNER JOIN atlantia a2 ON a2.time = Sub1.minTime AND a1.date = a2.date
 INNER JOIN atlantia a3 ON a3.time = Sub1.maxtime AND a1.date = a3.date
 GROUP BY a1.date , timeInterval

如果我想在不同的时间范围内聚合，即5分钟怎么办？我只用/ 5改变/ 60？

由于

Answer 1

我可以发现一些问题。

您正在将时间存储在角色字段中，当投射时，签名时间为10点20分被视为1020而不是10 * 60分钟+20分钟。因此当除以60时10:00是16而10:20是17时，因此在你的代码中它们是2个不同的时间间隔。

另一个问题是您已将时间作为字段返回，而不在GROUP BY子句中指定。它将返回的时间值来自未确定的行（通常是第一行但不总是）。可能最容易指定MIN（时间）。

SELECT 
    floor( (cast( SUBSTRING(time,1,2) AS SIGNED ) * 60 + cast( SUBSTRING(time,3,2) AS SIGNED )) /60 ) AS timeInterval, 
    date,
    MIN(time),
    MAX(high) AS high,
    MIN(low) as low,
    SUM(volume) as volume,
    (select 
            open
        from
            atlantia a2
        where
            a1.time = a2.time
        order by time
        limit 1) as open,
    (select 
            close
        from
            atlantia a2
        where
            a1.time = a2.time
        order by time desc
        limit 1) as close
FROM
    atlantia a1
GROUP BY date , timeInterval

可以清理sql以删除子查询。

修改

有一个游戏，这可能会这样做，但不确定效率，没有表我无法测试： -

SELECT 
    Sub1.timeInterval, 
    a1.date,
    MIN(a1.time),
    MAX(a1.high) AS high,
    MIN(a1.low) as low,
    SUM(a1.volume) as volume,
    MIN(a2.open) as open,
    MIN(a3.close) as close
FROM atlantia a1
INNER JOIN
(
    SELECT floor( (cast( SUBSTRING(time,1,2) AS SIGNED ) * 60 + cast( SUBSTRING(time,3,2) AS SIGNED )) /60 ) AS timeInterval, MIN(time) AS minTime, MAX(time) AS maxtime
    FROM atlantia
    GROUP BY timeInterval
) Sub1
ON floor( (cast( SUBSTRING(a1.time,1,2) AS SIGNED ) * 60 + cast( SUBSTRING(a1.time,3,2) AS SIGNED )) /60 ) = Sub1.timeInterval
INNER JOIN atlantia a2 ON a2.time = Sub1.minTime AND a1.date = a2.date
INNER JOIN atlantia a3 ON a3.time = Sub1.minTime AND a1.date = a3.date
GROUP BY a1.date , timeInterval