按日期分組，即使日期沒有條目

Question

我想要通過計算在同一天創建了多少個條目來可視化我的條目。

SELECT dayname(created_at), count(*) FROM logs 
group by day(created_at) 
ORDER BY created_at desc 
LIMIT 7 

所以我得到的是這樣的：

Thursday 4 
Wednesday 12 
Monday 4 
Sunday 1 
Saturday 20 
Friday 23 
Thursday 10 

但我也想有周二在那裏與0，所以我有一個星期。

有沒有辦法做到這一點與完整的MySQL或我需要更新結果之前，我可以把它給圖表？

編輯：

這是最終的查詢：

SELECT 
    DAYNAME(date_add(NOW(), interval days.id day)) AS day, 
    count(logs.id) AS amount 
FROM days LEFT OUTER JOIN 
    (SELECT * 
    FROM logs 
    WHERE TIMESTAMPDIFF(DAY,DATE(created_at),now()) < 7) logs 
    on datediff(created_at, NOW()) = days.id 
GROUP BY days.id 
ORDER BY days.id desc; 

表天包括從0到-6

Answer 1

你只需要補償的表，它可能是一個真正的表或一些像飛行一樣建造的東西，如select 0 ofs union all select -1 ...。

create table days (ofs int); 
insert into days (ofs) values 
    (0), (-1), (-2), (-3), 
    (-4), (-5), (-6), (-7); 

select 
    date_add('20160121', interval days.ofs day) as created_at, 
    count(data.id) as cnt 
from days left outer join logs data 
    on datediff(data.created_at, '20160121') = days.ofs 
group by days.ofs 
order by days.ofs; 

http://sqlfiddle.com/#!9/3e6bc7/1

對於性能，可能會更好，以限制數據（日誌）表中搜索：

select 
    date_add('20160121', interval days.ofs day) as created_at, 
    count(data.id) as cnt 
from days left outer join 
    (select * from logs where created_at between <start> and <end>) data 
    on datediff(data.created_at, '20160121') = days.offset 
group by days.offset 
order by days.offset; 

一個缺點是，你必須有一個參數化這個在幾個表達式中固定錨點日期。將真正日期的表格放在某個表格中可能會更好，因此您不必進行計算。

Answer 2

使用號RIGHT JOIN的日期表，所以您可以申請每一天和每一天的數據，無論是否有數據都有數據，簡單來說，精簡的日期將顯示爲CERO或NULL。

您可以創建日期表，某種日曆表。

id_day | day_date | 
-------------------- 
1  | 2016-01-01 | 
2  | 2016-01-02 | 
. 
. 
365 | 2016-12-31 | 

有了這張表，你可以涉及日期，然後提取日，月，周，任何你想要的MYSQL DATE AND TIME FUNCTIONS

SELECT t2.dayname(day_date), count(t1.created_at) FROM logs t1 right join dates_table t2 on t1.created_at=t2.day_date group by t2.day_date ORDER BY t1.created_at desc LIMIT 7