按时间汇总记录

Question

我有一个查询按时间执行聚合记录，但我有一些错误。

我的情况：我的股票行情数据每分钟5.6一个创纪录的0900存储在MySQL数据库VERS高达1730，这是511点的记录每天

我需要汇总这些数据进行不同的时间帧，让我们假设5分钟，使得

volume5min - >总和体积0900：0904

open5min - 为0900记录（在范围内首开值）

close5min>开值 - >最高每日0900时至价值 - 0904记录（在范围内最后收盘值）

high5min>收盘值0904范围

low5min - >最低在0900的值0904范围

等。

我有一个查询做到这一点，但我得到的开启和关闭聚合值误差

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 

这是我得到执行查询

打开和关闭是没有正确汇总，而其他列看起来不错。

更重要的是，改变总的时间框架，我得到不正确的时间，如下面

从9时开始的60分钟的例子，我应该有1000，1100等，而现在我也有1020，1140等

列类型有：

日期：CHAR 时间：CHAR 其余全部为双卷却是INTEGER。

如何修改此查询以正确汇总值？

编辑：，以验证您的最后一个查询，我已经手动检查任何一个小时，这些都是正确的价值观，我应该得到

查询返回的密切价值观的差异，所有其余它的确定

关闭值应为时限的最后一个记录，即时间betwee n 0900和0959 close是紧密列中的0959值。

编辑2：看来我发现其中的诀窍是，现在有了这个查询这一切工作正常

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框架聚集，即5分钟？我只是用/ 5更换两个/ 60？

谢谢

Answer 1

我可以发现几个问题。

您正在将时间存储在字符字段中，并且在将10:20点钟的符号进行投射时，将视为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，因此在你的代码中它们有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，而未在GROUP BY子句中指定它。它返回的时间值来自未确定的行（通常是第一个，但并非总是）。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 

可能会清理sql以删除子查询。

编辑

了发挥，这可能做到这一点，但不知道效率和不表，我不能测试： -

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