Select语句，该语句从sql中的时间戳计数匹配的日期（而不是时间）

Question

I don't want to combine or group them. I only want to count how many dates (not the hours/min/sec) match in a given table then use php to highlight the given days on a simple calendar. (Thus highlighting days that have more than two entries.)

I just can't figure out the statement. Here's an example of what I've tried. I pulls SOME rows, but it's not 100%.

    SELECT * 
    FROM `foo_table` 
    WHERE DAY(`start`) IN
    (
        SELECT DAY(`start`)         
        FROM `foo_table`
        WHERE MONTH(`start`) IN (01) 
        AND MONTH(`end`) IN (01) 
        GROUP BY `start` 
        HAVING COUNT(*) > 1 
    )
    ORDER BY `start`

Here's the table:

    id          start                      end
    1    2014-01-01 10:00:00       2014-01-02 10:00:00
    2    2014-01-02 10:00:00       2014-01-04 10:00:00
    3    2014-01-03 10:00:00       2014-01-06 10:00:00
    4    2014-02-01 10:00:00       2014-02-02 10:00:00
    5    2014-02-01 10:00:00       2014-02-02 10:00:00
    6    2014-10-01 10:00:00       2014-10-19 10:00:00

I want the statement to show for a query of: January 3 unique entries. February 2 matching entries. And so on.

I'm using php to change the font of a calendar when there is one, two, or three entries for a given day. I just can't seem to figure out how to find out the amount of entries and which days.

Answer 1

If I'm understanding you correctly, would something like this work for you:

SELECT DAY(`start`), COUNT(*) 
FROM foo_table 
WHERE DATE_FORMAT(`start`, "%Y-%m") = '2014-03' 
GROUP BY 1;

This would return you something similar to:

+-----------+----------+
|         1 |      266 |
|         2 |      260 |
|         3 |      304 |
|         4 |      298 |
|         5 |      251 |
|         6 |      293 |
|         7 |      310 |
|         8 |      243 |
|         9 |      248 |
|        10 |      288 |
|        11 |      290 |
|        12 |      245 |
|        13 |      315 |
|        14 |      333 |
|        15 |      298 |
|        16 |      268 |
|        17 |      282 |
|        18 |      245 |
|        19 |      252 |
|        20 |      293 |
|        21 |       63 |
+-----------+----------+

Answer 2

Your subquery needs to group by date(start) :

SELECT * 
FROM `foo_table` 
WHERE DATE(`start`) IN (SELECT DATE(`start`)         
                        FROM `foo_table`
                        WHERE MONTH(`start`) IN (01) AND MONTH(`end`) IN (01) 
                        GROUP BY DATE(`start`)
--------------------------------^
                        HAVING COUNT(*) > 1 
                      )
ORDER BY `start`;

From your description, though, it would seem that the output of the subquery would be sufficient because it identifies the days with multiple entries.