查找每天的平均值，如果当天没有记录，则取最新的平均值

Question

I'm not sure that the tile is correct nor specific..

I have a table that looks like this:

2013-01-01
a: 100

2013-01-02
b: 50

2013-01-03
c: 75
b: 200

Where "a, b, c" user ids.

I need a query that will return:

2013-01-01
100 / 1 = 100

2013-01-02
(100 + 50) / 2 = 75

2013-01-03
(100 + 200 + 75) / 3 = 125

[[2013-01-01, 100], [2013-01-02, 75], [2013-01-03, 125]]

It seems for a certein date I can do this with:

SELECT AVG(column1) FROM table a 
  JOIN (
   SELECT MAX(date) AS max_date, user_id FROM table 
     WHERE date < "2013-09-01" GROUP BY user_id) b 
   ON b.user_id = a.user_id AND a.date = b.max_date;

However, I have an interval which can be split by day, week, month or year: ['2013-01-01', '2013-01-02',... '2014-01-01'] (continuous)

Is there a faster way to do that instead of running 100 queries for a 100 day interval?

sqlfiddle: http://sqlfiddle.com/#!2/d74fa/2

Desired result: single query that will return

2013-01-01, 100
2013-01-02, 75
2013-01-03, 125

Answer 1

If you need the daily average you have to use:

SELECT AVG(value)
FROM readings r
GROUP BY(date);

I am not sure why you are doing the average up to the day in the analytical description.