Select results inclusively based on NOW()

Question

My query is the following:

SELECT count(db.user_data.page_url) AS total_urls, db.authors.author
FROM db.authors
LEFT JOIN db.user_data ON db.authors.author_id = db.user_data.id
WHERE (db.user_data.date > (now() - INTERVAL 30 MINUTE))
AND db.user_data.domain LIKE '%example.com%'
GROUP BY author;

It returns all results between now() and last 30 minutes, that's fine. But those results are based on period (interval) of 30 minutes starting from current moment (in seconds).

How to select all rows based on now() - INTERVAL 30 MINUTE ,but including all results from the last 30-th minute inclusively? Because now the interval is "moving" by current moment + 30 last minutes. So each new second I am getting different results, because of "sliding", "moving" interval?

Andriy M · Accepted Answer · 2018-06-19T10:57:07.840

4

Subtract the seconds part of NOW() from the value of NOW() and you will get the timestamp marking the beginning of the current minute. Now you can subtract your interval to get the beginning of the minute that was 30 minutes ago:

WHERE (date > (NOW() - INTERVAL SECOND(NOW()) SECOND - INTERVAL 30 MINUTE))

edited Jun 19 '18 at 10:57

answered Jun 18 '18 at 18:40

Andriy M

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score 2 · Answer 2 · answered Jun 18 '18 at 18:48

Another solution would be to change the date to timestamp, remove the reminder of dividing by 60 as follows:

select now();
+---------------------+
| now()               |
+---------------------+
| 2018-06-18 14:46:36 |
+---------------------+
1 row in set (0.00 sec)

select from_unixtime((unix_timestamp((now() - INTERVAL 30 MINUTE))div 60)*60) as 30_min_ago;
+---------------------+
| 30_min_ago          |
+---------------------+
| 2018-06-18 14:16:00 |
+---------------------+

Select results inclusively based on NOW()

2 Answers2