How to add value for every date in MySQL query - mysql

Question

I have following sql-query:
SELECT DATE(time), ROUND(AVG(out_temp),2)
FROM data_table
WHERE id= 1 AND time BETWEEN '2012-08-18' AND '2012-08-30'
GROUP BY DATE(time)
ORDER BY time ASC
This returns:
date avg_temp
2012-08-18 11.41
2012-08-19 5.90
2012-08-28 11.22
2012-08-29 10.07
Everything works well so far... but I would like to add missing dates with constant value like this:
date avg_temp
2012-08-18 11.41
2012-08-19 5.90
2012-08-20 <value>
... ...
2012-08-27 <value>
2012-08-28 11.22
2012-08-29 10.07
How should I modify my query? Could somebody help me with this problem? I read some posts about creating a separate calendar table with prefilled date values, but I still didnt get it to work.

Answer 1

If your data table actually has data on every date, you can do this:
SELECT thedate, coalesce(ROUND(AVG(out_temp),2), <value>)
FROM (select distinct date(time) as thedate
from data_table
) dates left outer join
data_table dt
on date(time) = thedates.date
WHERE id= 1 AND thedate BETWEEN '2012-08-18' AND '2012-08-30'
GROUP BY thedate
ORDER BY time ASC
What you need is a driving table to generate the dates that you need. You can then left join to this table, to get the summaries you want. The COALESCE function lets you put in your default value.

Answer 2

Create a table with all dates you need, and then do a LEFT JOIN. E.g.
CREATE TABLE calendar ( day DATE PRIMARY KEY );
Then insert into the table, probably with a loop on your programming language (pseudocode):
for day in day_range( start_date, end_date ):
query( 'INSERT INTO calendar VALUES ( ' + day + ' );' );
And then do a LEFT JOIN:
SELECT DATE(day), ROUND(AVG(out_temp),2)
FROM data_table LEFT JOIN calendar ON data_table.time = calendar.day
WHERE id= 1 AND day BETWEEN '2012-08-18' AND '2012-08-30'
GROUP BY DATE(day)
ORDER BY day ASC