一個月中每天的高峯時間表MySQL

Question

我正在嘗試進行查詢以獲得特定月份每天的高峯時間。 我的表看起來像這樣：

id  idproduct  created_at 
1  021354684  2011-10-01 20:25:48 
2  033546835  2011-10-01 20:30:15 
3  055965654  2011-10-01 20:45:20 
4  012975343  2011-10-02 14:03:36 
5  021354684  2011-10-02 15:55:48 
6  033546835  2011-10-02 16:30:15 
7  055965654  2011-10-02 16:45:20 
8  012975343  2011-10-02 18:53:36 
9  021354684  2011-10-03 08:55:48 
10  033546835  2011-10-03 09:30:15 
11  055965654  2011-10-03 14:03:20 
12  012975343  2011-10-03 14:03:36 

我試圖得到的是這樣的事情...：

day  rush_hour number_of_rows 
1  20:00  3 
2  16:00  5 
3  14:00  4 

是否有可能得到這樣的表？你們能幫我嗎？

我犯了一個錯誤，對此抱歉。行數應占總當天賣出，而不是在這一個小時:(不好意思項目

Answer 1

http://sqlfiddle.com/#!2/5b87b/7

幫助

首先，計算每一天的每一個小時的計數值（到視圖，因爲我們將在下面兩次使用）：

CREATE VIEW hours AS 
SELECT 
    DATE(created_at) AS d, 
    HOUR(created_at) AS h, 
    COUNT(*) AS c 
FROM item 
GROUP BY DATE(created_at), HOUR(created_at); 

最終查詢：

SELECT 
    hours.d AS `day`, 
    hours.h AS `rush_hour`, 
    hours.c AS `count` 

-- get the max count for every day 
FROM (
     SELECT 
     d,   -- the day 
     MAX(c) as c -- the count 
     FROM hours 
     GROUP BY d 
    ) AS maxc 

-- find the actual hour(s) with the max count for every day: 
INNER JOIN hours ON hours.c = maxc.c 
       AND hours.d = maxc.d; 

Answer 2

試試這個：

SELECT dayofyear(created_at) as day, hour(created_at) as rush_hour, count(*) as number_of_rows 
FROM table 
GROUP BY dayofyear(created_at), hour(created_at); 

Answer 3

你會想看看MySQL Date Functions，他們提供你一些與此

SELECT 
    day(created_at) as day, 
    hour(created_at) as rush_hour, 
    count(1) as num_rows 
FROM item 
GROUP BY 
    day(created_at), hour(created_at) 

http://sqlfiddle.com/#!2/62a15/2/0

Answer 4

這未做一個觀點：

SELECT ddd.day, eee.rush_hour, ddd.maxo 
FROM 
(select day, max(num_rows) as maxo from (
SELECT 
    day(created_at) as day, 
    hour(created_at) as rush_hour, 
    count(1) as num_rows 
FROM item 
GROUP BY 
    day(created_at), hour(created_at) 
) as groupo group by day) as ddd 

LEFT JOIN 
(SELECT 
    day(created_at) as day, 
    hour(created_at) as rush_hour, 
    count(1) as num_rows 
FROM item 
GROUP BY 
    day(created_at), hour(created_at) 
) as eee on ddd.day=eee.day and ddd.maxo=eee.num_rows 

我可以想象它被格式化得更好或者有更多相關的別名，但是這裏有很多子選擇。

並感謝SQLfiddlers將數據放在那裏。

而且我認爲，如果你有兩個小時並列最高數量的數字，它們都會顯示出來，所以你會得到兩個（或更多）記錄。