Grouping MySQL results by 7 day increments - mysql

Question

Hoping someone might be able to assist me with this.
Assume I have the table listed below. Hosts can show up multiple times on the same date, usually with different backupsizes.
+------------------+--------------+
| Field | Type |
+------------------+--------------+
| startdate | date |
| host | varchar(255) |
| backupsize | float(6,2) |
+------------------+--------------+
How could I find the sum total of backupsize for 7 day increments starting with the earliest date, through the last date? I don't mind if the last few days get cut off because they don't fall into a 7 day increment.
Desired output (prefered):
+------------+----------+----------+----------+-----
|Week of | system01 | system02 | system03 | ...
+------------+----------+----------+----------+-----
| 2014/07/30 | 2343.23 | 232.34 | 989.34 |
+------------+----------+----------+----------+-----
| 2014/08/06 | 2334.7 | 874.13 | 234.90 |
+------------+----------+----------+----------+-----
| ... | ... | ... | ... |
OR
+------------+------------+------------+------
|Host | 2014/07/30 | 2014/08/06 | ...
+------------+------------+------------+------
| system01 | 2343.23 | 2334.7 | ...
+------------+------------+------------+-------
| system02 | 232.34 | 874.13 | ...
+------------+------------+------------+-------
| system03 | 989.34 | 234.90 | ...
+------------+------------+------------+-------
| ... | ... | ... |
Date format is not a concern, just as long as it gets identified somehow. Also, the order of the hosts is not a concern either. Thanks!

Answer 1

The simplest way is to get the earliest date and just count the number of days:
select x.minsd + interval floor(datediff(x.minsd, lb.startdate) / 7) day as `Week of`,
host,
sum(backupsize)
from listedbelow lb cross join
(select min(startdate) as minsd from listedbelow lb) x
group by floor(datediff(x.minsd, lb.startdate) / 7)
order by 1;
This produces a form with week of and host on each row. You can pivot the results as you see fit.

Answer 2

I'll assume that what you want is the sum of bakcupsize grouped by host and that seven-day interval you are talking about.
My solution would be something like this:
You need to define the first date, and then "create" a column with the date you want (the end of the seven-day period)
Then I would group it.
I think temporary tables and little tricks with temp variables are the best way to tackle this, so:
drop table if exists temp_data;
create temporary table temp_data
select a.*
-- The #d variable will have the date that you'll use later to group the data.
, #d := case
-- If the current "host" value is the same as the previous one, then...
when #host_prev = host then
-- ... if #d is not null and is within the seven-day period,
-- then leave the value of #d intact; in other case, add 7 days to it.
case
when #d is not null or a.startdate <= #d then #d
-- The coalesce() function will return the first not null argument
-- (just as a precaution)
else dateadd(coalesce(#d, a.startdate), interval +7 day)
end
-- If the current "host" value is not the same as the previous one,
-- then take the current date (the first date of the "new" host) and add
-- seven days to it.
else #d = dateadd(a.startdate, interval +7 day)
end as date_group
-- This is needed to perform the comparisson in the "case" piece above
, #host_prev := a.host as host2
from
(select #host_prev = '', #d = null) as init -- Initialize the variables
, yourtable as a
-- IMPORTANT: This will only work if you order the data properly
order by a.host, a.startdate;
-- Add indexes to the temp table, to make things faster
alter table temp_data
add index h(host),
add index dg(date_group)
-- OPTIONAL: You can drop the "host2" column (it is no longer needed)
-- , drop column host2
;
Now, you can get the grouped data:
select a.host, a.date_group, sum(a.bakcupsize) as backupsize
from temp_data as a
group by a.host, a.date_group;
This will give you the unpivoted data. If you want to build a pivot table with it, I recommend you take a look to this article, and/or read this question and its answers. In short, you'll have to build a "dynamic" sql instruction, prepare a statement with it and execute it.
Of course, if you want to group this by week, there's a simpler approach:
drop table if exists temp_data2;
create temporary table temp_data2
select a.*
-- The following will give you the end-of-week date
, dateadd(a.startdate, interval +(6 - weekday(a.startdate)) day) as group_date
from yourtable as a;
alter table temp_data
add index h(host),
add index dg(date_group);
select a.host, a.date_group, sum(a.bakcupsize) as backupsize
from temp_data as a
group by a.host, a.date_group;
I leave the pivot part to you.

Answer 3

So I was able to determine a solution that fit my needs using a procedure I created by putting together concepts from your recommended solutions as well as some other other solutions I found on this site. The procedure SUM's by 7 day increments as well as does a pivot.
DELIMITER $$
CREATE PROCEDURE `weekly_capacity_by_host`()
BEGIN
SELECT MIN(startdate) into #start_date FROM testtable;
SET #SQL = NULL;
SELECT
GROUP_CONCAT(DISTINCT
CONCAT(
'SUM(if(host=''',host,''', backupsize, 0)) as ''',host,''''
)
) INTO #SQL
FROM testtable;
SET #SQL = CONCAT('SELECT 1 + DATEDIFF(startdate, ''',#start_date,''') DIV 7 AS week_num
, ''',#start_date,''' + INTERVAL (DATEDIFF(startdate, ''',#start_date,''') DIV 7) WEEK AS week_start,
', #SQL,'
FROM testtable group by week_num'
);
PREPARE stmt FROM #SQL;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
END$$
DELIMITER ;
Output appears as follows:
mysql> call weekly_capacity_by_host;
+----------+------------+----------+----------+----------+----------+
| week_num | week_start | server01 | server02 | server03 | server04 |
+----------+------------+----------+----------+----------+----------+
| 1 | 2014-06-11 | 1231.08 | 37.30 | 12.04 | 68.17 |
| 2 | 2014-06-18 | 1230.98 | 37.30 | 11.76 | 68.13 |
| 3 | 2014-06-25 | 1243.12 | 37.30 | 8.85 | 68.59 |
| 4 | 2014-07-02 | 1234.73 | 37.30 | 11.77 | 67.80 |
| 5 | 2014-07-09 | 341.32 | 0.04 | 0.14 | 4.94 |
+----------+------------+----------+----------+----------+----------+
5 rows in set (0.03 sec)