简体   繁体   中英

MySQL to get the count of rows that fall on a date for each day of a month

I have a table that contains a list of community events with columns for the days the event starts and ends. If the end date is 0 then the event occurs only on the start day. I have a query that returns the number of events happening on any given day:

SELECT COUNT(*) FROM p_community e WHERE 
    (TO_DAYS(e.date_ends)=0 AND DATE(e.date_starts)=DATE('2009-05-13')) OR
    (DATE('2009-05-13')>=DATE(e.date_starts) AND DATE('2009-05-13')<=DATE(e.date_ends))

I just sub in any date I want to test for "2009-05-13".

I need to be be able to fetch this data for every day in an entire month. I could just run the query against each day one at a time, but I'd rather run one query that can give me the entire month at once. Does anyone have any suggestions on how I might do that?

And no, I can't use a stored procedure.

Try:

SELECT COUNT(*), DATE(date) FROM table WHERE DATE(dtCreatedAt) >= DATE('2009-03-01') AND DATE(dtCreatedAt) <= DATE('2009-03-10') GROUP BY DATE(date);

This would get the amount for each day in may 2009.

UPDATED: Now works on a range of dates spanning months/years.

Unfortunately, MySQL lacks a way to generate a rowset of given number of rows.

You can create a helper table:

CREATE TABLE t_day (day INT NOT NULL PRIMARY KEY)

INSERT
INTO    t_day (day)
VALUES  (1),
        (2),
        …,
        (31)

and use it in a JOIN :

SELECT  day, COUNT(*)
FROM    t_day
JOIN    p_community e
ON      day BETWEEN DATE(e.start) AND IF(DATE(e.end), DATE(e.end), DATE(e.start))
GROUP BY
        day

Or you may use an ugly subquery:

SELECT  day, COUNT(*)
FROM    (
        SELECT  1 AS day
        UNION ALL
        SELECT  2 AS day
        …
        UNION ALL
        SELECT  31 AS day
        ) t_day
JOIN    p_community e
ON      day BETWEEN DATE(e.start) AND IF(DATE(e.end), DATE(e.end), DATE(e.start))
GROUP BY
        day

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM