使用 MySQL 按分鍾分組和計數

Question

我想從大型時間戳數據庫中獲取一天中每一分鍾的計數。 我希望能夠根據數據構建圖表。

那么一天有 1440 分鍾，我如何計算每分鍾的計數？

我可以在一小時內獲得每分鍾的計數，我對如何擴展以獲取一天中每一分鍾的計數感到困惑。

我有我目前正在使用的代碼，

 SELECT FROM_UNIXTIME(
         CEILING(UNIX_TIMESTAMP(`timestamp`)/900)*900
                    ) AS timeslice
     , COUNT(*) AS mycount
  FROM visitor
 WHERE `timestamp` >= '20012-01-01'
   AND `timestamp`  < '20012-02-14'
GROUP BY timeslice 

我希望這是有道理的，並感謝您的幫助！

有任何問題請問我

擔

Answer 1

根據您可以嘗試使用以下查詢的記錄數（檢查其執行時間）。

select d.everyminute, 
      (select count(*) from visitor 
              where `timestamp` between d.everyminute and 
                                        d.everyminute + interval 59 second) cnt 
      from 
(
 select @rownum:=@rownum+1, 
       date('2012-02-15') + interval (@rownum-1) minute everyminute from

(select 0 union all select 1 union all select 2) t,

(select 0 union all select 1 union all select 2 union all select 3 union all 
 select 4 union all select 5 union all select 6 union all select 7 union all 
 select 8 union all select 9) t1,

(select 0 union all select 1 union all select 2 union all select 3 union all 
 select 4 union all select 5 union all select 6 union all select 7 union all 
 select 8 union all select 9) t2,

(select 0 union all select 1 union all select 2 union all select 3 union all 
 select 4 union all select 5 union all select 6 union all select 7 union all 
 select 8 union all select 9) t3,

(select @rownum:=0) r where @rownum < 24*60
) d

結果是（我在測試數據庫中放入了一些時間戳）

2012-02-15 00:00:00 0
2012-02-15 00:01:00 0
2012-02-15 00:02:00 2
2012-02-15 00:03:00 1
2012-02-15 00:04:00 0
2012-02-15 00:05:00 1
2012-02-15 00:06:00 0
2012-02-15 00:07:00 0
2012-02-15 00:08:00 0
2012-02-15 00:09:00 0
2012-02-15 00:10:00 0

等等直到2012-02-15 23:59:00

現有記錄的時間戳為2012-02-15 00:02:00 、 2012-02-15 00:02:00 、 2012-02-15 00:03:00和2012-02-15 00:05:00

ps：考慮到夏季/冬季時間跳躍，您可以將where @rownum < 24*60重寫為having everyminute < (date('2012-02-15') + interval 1 day - interval 1 minute)

如您所見，即使在這一分鍾內 db 中沒有記錄，它也會每分鍾計數一次。

Answer 2

最簡單的方法是創建一個已包含分鍾的實用程序表。 如果你很聰明，你也會把小時和子午線。

您可以通過搜索數據倉庫時間維度找到更多信息。

Answer 3

看起來您不希望它按天/月/年分組，只需幾分鍾：

select minute, sum(if(adate is null, 0, 1)) DateAmount from (
  select @num := @num + 1 as minute from
    (select 1  union all select 2  union all select 3  union all
     select 4  union all select 5  union all select 6  union all
     select 7  union all select 8  union all select 9  union all
     select 10 union all select 11 union all select 12) as t1,
    (select 1  union all select 2  union all select 3  union all
     select 4  union all select 5  union all select 6  union all
     select 7  union all select 8  union all select 9  union all
     select 10 union all select 11 union all select 12) as t2,
    (select 1  union all select 2  union all select 3  union all
     select 4  union all select 5  union all select 6  union all
     select 7  union all select 8  union all select 9  union all
     select 10) as t3,
    (select @num := - 1) as t4
) as MinuteInDay
left join visitors v
on minute(v.aDate) = MinuteInDay.minute
group by minute

對於這組數據：

+---------------------+
|        aDate        |
+---------------------+
| 2012-01-01 00:00:00 |
| 2012-01-01 00:01:00 |
| 2012-01-01 00:02:00 |
| 2012-01-01 00:02:00 |
| 2012-01-02 00:02:00 |
| 2012-01-03 00:03:00 |
+---------------------+

這將導致：

+--------+------------+
| MINUTE | DATEAMOUNT |
+--------+------------+
| 0      | 1          |
| 1      | 1          |
| 2      | 3          |
| 3      | 1          |
| 4      | 0          |
| 5      | 0          |
| ...    | ...        |
| 1439   | 0          |
+--------+------------+

希望這可以幫助。

PS：如果您不需要 0 中的分鍾並且只忽略它們會容易得多！

Answer 4

我會做這樣的事情：

SELECT MOD(FLOOR(UNIX_TIMESTAMP( timestamp )/60), 1440) AS timeslice, COUNT(*) AS mycount FROM visitor WHERE時間戳>= '20012-01-01' AND時間戳< '20012-02-14' GROUP BY timeslice

它將計算自 unix 時代以來 1440 分鍾的批次，忽略夏令時等。