问题
I have a query that groups all entries from a table and groups them by the datetime column. This is all working great:
SELECT SUM( `value` ) AS `sum` , DATE(`datetime`) AS `dt``
FROM `entry`
WHERE entryid = 85
AND DATETIME BETWEEN '2010-01-01' AND '2010-03-01'
GROUP BY `dt`
ORDER BY `datetime`
The problem is, I need it to return a row even if nothing is found, with the value set to 0. I assume there's some MYSQL function that'll take care of this but can't seem to find it.
Thanks!
回答1:
MySQL doesn't have recursive functionality, so you're left with using the NUMBERS table trick -
Create a table that only holds incrementing numbers - easy to do using an auto_increment:
DROP TABLE IF EXISTS `example`.`numbers`; CREATE TABLE `example`.`numbers` ( `id` int(10) unsigned NOT NULL auto_increment, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Populate the table using:
INSERT INTO NUMBERS (id) VALUES (NULL)
...for as many values as you need.
Use DATE_ADD to construct a list of dates, increasing the days based on the NUMBERS.id value. Replace "2010-01-01" and "2010-03-01" with your respective start and end dates (but use the same format, YYYY-MM-DD) -
SELECT x.* FROM (SELECT DATE_ADD('2010-01-01', INTERVAL n.id - 1 DAY) FROM numbers n WHERE DATE_ADD('2010-01-01', INTERVAL n.id -1 DAY) <= '2010-03-01' ) x
LEFT JOIN onto your table of data based on the datetime portion:
SELECT DATE(x.dt) AS dt, COALESCE(SUM(e.value), 0) AS sum_value FROM (SELECT DATE_ADD('2010-01-01', INTERVAL n.id - 1 DAY) AS dt FROM numbers n WHERE DATE_ADD('2010-01-01', INTERVAL n.id -1 DAY) <= '2010-03-01' ) x LEFT JOIN ENTRY e ON DATE(e.datetime) = x.dt AND e.entryid = 85 GROUP BY DATE(x.dt)
Why Numbers, not Dates?
Simple - dates can be generated based on the number, like in the example I provided. It also means using a single table, vs say one per data type.
来源:https://stackoverflow.com/questions/3640151/grouping-by-date-return-row-even-if-no-records-found