按時間彙總記錄

Question

我有一個查詢按時間執行聚合記錄，但我有一些錯誤。

我的情況：我的股票行情數據每分鐘5.6一個創紀錄的0900存儲在MySQL數據庫VERS高達1730，這是511點的記錄每天

我需要彙總這些數據進行不同的時間幀，讓我們假設5分鐘，使得

volume5min - >總和體積0900：0904

open5min - 爲0900記錄（在範圍內首開值）

close5min>開值 - >最高每日0900時至價值 - 0904記錄（在範圍內最後收盤值）

high5min>收盤值0904範圍

low5min - >最低在0900的值0904範圍

等。

我有一個查詢做到這一點，但我得到的開啓和關閉聚合值誤差

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 

這是我得到執行查詢

打開和關閉是沒有正確彙總，而其他列看起來不錯。

更重要的是，改變總的時間框架，我得到不正確的時間，如下面

從9時開始的60分鐘的例子，我應該有1000，1100等，而現在我也有1020，1140等

列類型有：

日期：CHAR 時間：CHAR 其餘全部爲雙卷卻是INTEGER。

如何修改此查詢以正確彙總值？

編輯：，以驗證您的最後一個查詢，我已經手動檢查任何一個小時，這些都是正確的價值觀，我應該得到

查詢返回的密切價值觀的差異，所有其餘它的確定

關閉值應爲時限的最後一個記錄，即時間betwee n 0900和0959 close是緊密列中的0959值。

編輯2：看來我發現其中的訣竅是，現在有了這個查詢這一切工作正常

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 

如果我想在不同的時間框架聚集，即5分鐘？我只是用/ 5更換兩個/ 60？

謝謝

Answer 1

我可以發現幾個問題。

您正在將時間存儲在字符字段中，並且在將10:20點鐘的符號進行投射時，將視爲1020而不是10 * 60分鐘+20分鐘。因此，當除以60時，10:00是16，而10:20是17，因此在你的代碼中它們有2個不同的時間間隔。

另一個問題是您已將時間作爲字段返回，而未在GROUP BY子句中指定它。它返回的時間值來自未確定的行（通常是第一個，但並非總是）。可能最容易指定MIN（時間）。

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 

可能會清理sql以刪除子查詢。

編輯

了發揮，這可能做到這一點，但不知道效率和不表，我不能測試： -

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