MySQL查询中的三个小时间隔 [英] Three Hour Time Intervals in MySQL query
问题描述
在一个上一个问题上，可以创建一个表并将其填充为每月的几天，但我希望该表的填充略有不同:每月的每一天应具有三个不同的小时间隔.
On a previous question a table could be created and populated with the days of the month, but I'd like that table to be populated slightly different: each day of the month should have three different hour intervals.
根据该问题，此代码由 Tom Mac :
According to that question, this code by Tom Mac:
create table all_date
(id int unsigned not null primary key auto_increment,
a_date date not null,
last_modified timestamp not null default current_timestamp on update current_timestamp,
unique key `all_date_uidx1` (a_date));
然后
DELIMITER //
CREATE PROCEDURE populate_all_dates(IN from_date DATE, IN days_into_future INT)
BEGIN
DECLARE v_date DATE;
DECLARE ix int;
SET ix := 0;
SET v_date := from_date;
WHILE v_date <= (from_date + interval days_into_future day) DO
insert into all_date (a_date) values (v_date)
on duplicate key update last_modified = now();
set ix := ix +1;
set v_date := from_date + interval ix day;
END WHILE;
END//
DELIMITER ;
然后您可以运行:
call populate_all_dates('2011-10-01',30);
要填充十月的所有日期(或更改月份，请更改该函数的值)
To populate all dates for October (or whatever month, change the values of the function)
这样我可以运行以下查询
With that I could run the following query
select day(a.a_date) as 'October',
IFNULL(t.a1,0) as 'Auth1',
IFNULL(t.a2,0) as 'Auth2',
IFNULL(t.a50,0) as 'Auth50'
from all_date a
LEFT OUTER JOIN
(
SELECT date(wp.post_date) as post_date,
sum(case when wp.post_author = '1' then 1 else 0 end) as a1,
sum(case when wp.post_author = '2' then 1 else 0 end) as a2,
sum(case when wp.post_author = '50' then 1 else 0 end) as a50,
count(*) as 'All Auths'
FROM wp_posts wp
WHERE wp.post_type = 'post'
AND wp.post_date between '2011-10-01' and '2011-10-31 23:59:59'
GROUP BY date(wp.post_date)
) t
ON a.a_date = t.post_date
where a.a_date between '2011-10-01' and '2011-10-31'
group by day(a.a_date);
然后我将得到一个表格，其中包含按作者和日期排列的WordPress博客中的帖子数，类似于:
And I would get a table with the number of posts in my WordPress blog by author and day, similar to this:
+---------+---------+-------+------+---------+
| October | Auth1 | Auth2 | Auth3| Auth4 |
+---------+---------+-------+------+---------+
| 1 | 0 | 0 | 0 | 0 |
| 2 | 0 | 0 | 1 | 0 |
| 3 | 4 | 4 | 6 | 2 |
| 4 | 4 | 3 | 5 | 2 |
| 5 | 7 | 0 | 5 | 2 |
| 6 | 4 | 4 | 0 | 2 |
| 7 | 0 | 2 | 1 | 2 |
| 8 | 0 | 0 | 7 | 0 |
.....
etc
但是我想要的是每天分成三行，每行分别对应以下时间范围:
But what I'dlike to have is each day divided in three different rows, each one corresponding to the following time ranges:
00:00-14:30 14:31-18:15 18:16-23:59
00:00-14:30 14:31-18:15 18:16-23:59
因此，该表应显示类似的内容(例如，我不知道如何显示每个时间范围，因此一个好的方法应该是第1天，时间范围1(1-1)等).
So the table should show something like (for example, I don't know how each of the time ranges could be shown, so a good way should be day 1, time range 1 (1-1), etc).
+---------+---------+-------+------+---------+
| October | Auth1 | Auth2 | Auth3| Auth4 |
+---------+---------+-------+------+---------+
| 1-1 | 0 | 0 | 0 | 0 |
| 1-2 | 0 | 0 | 0 | 0 |
| 1-3 | 0 | 0 | 0 | 0 |
| 2-1 | 0 | 0 | 1 | 0 |
| 2-2 | 0 | 0 | 0 | 0 |
| 2-3 | 0 | 0 | 0 | 0 |
| 3-1 | 1 | 2 | 3 | 0 |
| 3-2 | 1 | 2 | 2 | 2 |
| 3-3 | 2 | 0 | 1 | 0 |
etc...
如您所见，三行总和等于当天的前一个唯一行.
As you can see, the three rows sum is equivalent to each of the previous unique row for the day.
有可能吗?
推荐答案
使用(UPDATE#2)
use (UPDATE #2)
SELECT
a.a_datetm as 'October',
IFNULL(p.a1,0) as 'Auth1',
IFNULL(p.a2,0) as 'Auth2',
IFNULL(p.a50,0) as 'Auth50'
FROM
(
SELECT CONCAT (day(X.a_date), '-1') AS a_datetm
FROM all_date X
WHERE X.a_date between '2011-10-01' and '2011-10-31'
UNION ALL
SELECT CONCAT (day(Y.a_date), '-2') AS a_datetm
FROM all_date Y
WHERE Y.a_date between '2011-10-01' and '2011-10-31'
UNION ALL
SELECT CONCAT (day(Z.a_date), '-3') AS a_datetm
FROM all_date Z
WHERE Z.a_date between '2011-10-01' and '2011-10-31'
) a
LEFT OUTER JOIN
(
SELECT
CONCAT (day(wp.post_date), (CASE WHEN (TIME(wp.post_date) < '14:31:00') THEN '-1' WHEN (TIME(wp.post_date) BETWEEN '14:31:00' AND '18:15:59') THEN '-2' ELSE '-3' END )) AS a_datetm,
sum(case when wp.post_author = '1' then 1 else 0 end) as a1,
sum(case when wp.post_author = '2' then 1 else 0 end) as a2,
sum(case when wp.post_author = '50' then 1 else 0 end) as a50,
count(*) as 'All Auths'
FROM wp_posts wp
WHERE wp.post_type = 'post'
AND wp.post_date between '2011-10-01' and '2011-10-31 23:59:59'
GROUP BY CONCAT (day(wp.post_date), (CASE WHEN (TIME(wp.post_date) < '14:31:00') THEN '-1' WHEN (TIME(wp.post_date) BETWEEN '14:31:00' AND '18:15:59') THEN '-2' ELSE '-3' END ))
) p
ON a.a_datetm = p.a_datetm
ORDER BY a.a_datetm ASC;
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