如何从小时数据库中存储日期值的表中获取每日总数的平均值和标准偏差？

Question

I a do have a table that stores a total count of error by hour, by day.

I need to calculate the average of the total number of errors by day, I was trying something like this:

SELECT 
  error_code, 
  AVG(total) as avg, 
  STDDEV_SAMP(total) as std
FROM 
  error_monitoring 
WHERE
  date_sub(curdate(), interval 30 day) <= date 
GROUP BY
  error_code 

But the result of the query is based on the hourly values I have in the table and no the daily totals.

The table in database shows as follow:

id | error_code | total | average | standard_deviation | date 
1  | W0334      | 2131  | 81      | 163.349            | 2016-12-20 23:00:00
2  | W0096      | 910   | 45      | 132.915            | 2016-12-20 15:00:00
3  | W0334      | 120   | 81      | 163.349            | 2016-12-20 08:00:00

So, how I can calculate this based on daily totals?

Answer 1

If you want the "average by day", you can calculate the total and then divide by the number of days:

SELECT error_code, SUM(total) / COUNT(DISTINCT date(date)) as avg
FROM error_monitoring
WHERE date_sub(curdate(), interval 30 day) <= date
GROUP BY error_code 

Your code also includes STDDEV_SAMP() . If you also want that calculation, then you would need to use a subquery, aggregating by day. Assuming that all error codes are available on each day:

SELECT error_code, AVG(day_total), STDDEV_SAMP(day_total)
FROM (SELECT error_code, date(date) as dte, SUM(total) as day_total 
      FROM error_monitoring
      WHERE date_sub(curdate(), interval 30 day) <= date
      GROUP BY error_code, date(date)
     ) em
GROUP BY error_code;