I have a table:
idint(11) NOT NULL
type varchar(64) NOT NULL
created_on datetime NOT NULL
Can I count all type devices (iSO, Android...) by week in one query when use doctrine2?
Sample output:
Week iOS Android Other
1 4 5 6
2 3 5 9
This is entirely possible, using methods similar to that detailed in Output Sum of some column in week intervals throughout a year, week dates consistent with day however not easily in the output you described. The best way to handle this will be to do a SELECT, grouping on WEEK and TYPE. I.e.
SELECT
-- Get the week and month of each row
YEAR(created_on) AS Year,
WEEK(created_on) AS Week,
type
-- Count how many rows are in this group
COUNT(*) AS frequency
FROM yourTable AS yt
-- Order by month and then week, so that week 4 in Jan comes before week 4 in Feb
GROUP BY
Year ASC,
Week ASC,
type ASC
This would give output such as
Year Week Type frequency
2013 1 'iOS' 4
2013 1 'Android' 5
2013 1 'Other' 6
2013 2 'iOS' 3
2013 2 'Android' 5
2013 2 'Other' 9
Related
I have a MySQL requirement to select data from a table based on a start date and end date and group it by weekly also selecting the data in reverse order by date. Assume that, I have chosen the start date as 1st November and the end date as 04 December. Now, I would like to fetch the data as 04 December to 28 November, 27 November to 20 November, 19 November to 12 November and so on and sum the value count for that week.
Given an example table,
id
value
created_at
1
10
2021-10-11
2
13
2021-10-17
3
11
2021-10-25
4
8
2021-11-01
5
1
2021-11-10
6
4
2021-11-18
7
34
2021-11-25
8
17
2021-12-04
Now the result should be like 2021-12-04 to 2021-11-28 as one week, following the same in reverse order and summing the column value data for that week. I have tried in the query to add the interval of 7 days after the end date but it didn't work.
SELECT count(value) AS total, MIN(R.created_at)
FROM data_table AS D
WHERE D.created_at BETWEEN '2021-11-01' AND '2021-12-04' - INTERVAL 7 DAY ORDER BY D.created_at;
And it's also possible to have the last week may have lesser than 7 days.
Expected output:
end_interval
start_interval
total
2021-12-04
2021-11-27
17
2021-11-27
2021-11-20
34
2021-11-20
2021-11-13
4
2021-11-13
2021-11-06
1
2021-11-06
2021-10-30
8
2021-10-30
2021-10-25
11
Note that the last week is only 5 days depending upon the selected from and end dates.
One option to address this problem is to
generate a calendar of all your intervals, beginning from last date till first date, with a split of your choice, using a recursive query
joining back the calendar with the original table
capping start_interval at your start_date value
aggregating values for each interval
You can have three variables to be set, to customize your date intervals and position:
SET #start_date = DATE('2021-10-25');
SET #end_date = DATE('2021-12-04');
SET #interval_days = 7;
Then use the following query, as already described:
WITH RECURSIVE cte AS (
SELECT #end_date AS end_interval,
DATE_SUB(#end_date, INTERVAL #interval_days DAY) AS start_interval
UNION ALL
SELECT start_interval AS end_interval,
GREATEST(DATE(#start_date), DATE_SUB(start_interval, INTERVAL #interval_days DAY)) AS start_interval
FROM cte
WHERE start_interval > #start_date
)
SELECT end_interval, start_interval, SUM(_value) AS total
FROM cte
LEFT JOIN tab
ON tab.created_at BETWEEN start_interval AND end_interval
GROUP BY end_interval, start_interval
Check the demo here.
I have list of records in table A and want to get the count of records for each month in range of days, starting from first of this month to the current day. So if we have 3 months and the current day is 19, So for the first month i need count of records between 1st to 19th of the month, and for the second month from 1st to 19th and so on.
id time
1 2005-07-05 14:10:29
2 2005-07-12 15:47:35
3 2005-08-02 16:38:53
4 2005-08-04 10:48:12
5 2006-08-22 17:34:28
6 2006-09-01 22:11:35
7 2006-09-09 15:10:19
8 2006-09-06 21:55:56
The desired Result:
time count
2005-07 2
2005-08 2
2006-09 3
we ignored the record id 5 because we just count the days between 1st and 19th for each month.
How can i do this?
You cam filter the rows with the function DAY() and aggregate:
SELECT DATE_FORMAT(time, '%Y-%m') yearmonth,
COUNT(*) counter
FROM tablename
WHERE DAY(time) <= DAY(CURRENT_DATE)
GROUP BY yearmonth
See the demo.
Results:
yearmonth
counter
2005-07
2
2005-08
2
2006-09
3
I have tried looking at some similar examples like group by date range and weekdays etc but I couldnt fix it on my query.
as per my sample data screenshot, I need to only return
sum(salesamount)/sum(salescount) for week 1
and
sum(salesamount)/sum(salescount) for week 2.
Each of the week contain 5 days (in this example is wednesday - sunday).
My Attempt:
select salesstartdate, date_add(salesstartdate, interval 5 day) as gdate,
salesamount, salescount, sum(salesamount)/sum(salescount) as ATV
from testing
group by gdate;
My desired output is:
Week 1 15.34173913
Week 2 15.80365088
Calculation to get week 1 is (3507.1+3639.97+5258.77+8417.04+5994.48)/(285+273+344+478+368)
Calculation to get week 2 is the same as above except the date would now be from 8 to 12 of June.
You can do it with a subquery. In order to first group your result set properly and then execute aggregation on it:
SELECT
concat('WEEK', ' ', weekno) as `Week #`,
MIN(salesstartdate) as startDate,
MAX(salesstartdate) as endDate,
sum(salesamount)/sum(salescount) as ATV
FROM
(
SELECT
salesstartdate,
salesamount,
salescount,
WEEKOFYEAR(salesstartdate) as weekno -- get the week number of the current year
FROM
weekno
WHERE
WEEKDAY(salesstartdate) BETWEEN 2 AND 6 -- get index of week day
) as weeks
GROUP BY
weekno
I have used 2 MySQL functions here:
WEEKOFYEAR()
WEEKDAY()
Output:
WEEK 23 | 2016-06-08 | 2016-06-12 | 15.8040
WEEK 24 | 2016-06-16 | 2016-06-19 | 15.9323
and without subquery as well:
SELECT
concat('WEEK', ' ', WEEKOFYEAR(salesstartdate)) as `Week #`,
MIN(salesstartdate) as startDate,
MAX(salesstartdate) as endDate,
sum(salesamount)/sum(salescount) as ATV
FROM
weekno
WHERE
WEEKDAY(salesstartdate) BETWEEN 2 AND 6 -- get index of week day
GROUP BY
WEEKOFYEAR(salesstartdate)
You can do this way
select SUBDATE(salesstartdate, WEEKDAY(salesstartdate)) as week_range
, sum(salesamount)/sum(salescount)
from testing
where salesstartdate between SUBDATE(salesstartdate, WEEKDAY(salesstartdate))
and date_add(SUBDATE(salesstartdate, WEEKDAY(salesstartdate)), interval 5 day))
Group by week_range
I have a table with a column named timestamp
i want to be able to order by month(timestamp) and year(timestamp) and group the months and years together
so for example, if i had the following timestamps:
2014-01-01
2014-02-01
2014-05-01
2015-01-01
i want to show in this order
MONTH YEAR
1 2015
5 2014
2 2014
1 2014
You can pick month and year from timestamp with MONTH and YEAR:
GROUP BY MONTH(field_with_ts) , YEAR(field_with_ts)
and the same thing with the ORDER BY clause.
Try this:
substring(date, 8 , 2) as Month, substring(date, 1, 3) as Year
See, if that works.
I am having an issue with a SELECT command in MySQL. I have a database of securities exchanged daily with maturity from 1 to 1000 days (>1 mio rows). I would like to get the outstanding amount per day (and possibly per category). To give an example, suppose this is my initial dataset:
DATE VALUE MATURITY
1 10 3
1 15 2
2 10 1
3 5 1
I would like to get the following output
DATE OUTSTANDING_AMOUNT
1 25
2 35
3 15
Outstanding amount is calculated as the total of securities exchanged still 'alive'. That means, in day 2 there is a new exchange for 10 and two old exchanges (10 and 15) still outstanding as their maturity is longer than one day, for a total outstanding amount of 35 on day 2. In day 3 instead there is a new exchange for 5 and an old exchange from day 1 of 10. That is, 15 of outstanding amount.
Here's a more visual explanation:
Monday Tuesday Wednesday
10 10 10 (Day 1, Value 10, matures in 3 days)
15 15 (Day 1, 15, 2 days)
10 (Day 2, 10, 1 day)
5 (Day 3, 5, 3 days with remainder not shown)
-------------------------------------
25 35 15 (Outstanding amount on each day)
Is there a simple way to get this result?
First of all in the main subquery we find SUM of all Values for current date. Then add to them values from previous dates according their MATURITY (the second subquery).
SQLFiddle demo
select T1.Date,T1.SumValue+
IFNULL((select SUM(VALUE)
from T
where
T1.Date between
T.Date+1 and T.Date+Maturity-1 )
,0)
FROM
(
select Date,
sum(Value) as SumValue
from T
group by Date
) T1
order by DATE
I'm not sure if this is what you are looking for, perhaps if you give more detail
select
DATE
,sum(VALUE) as OUTSTANDING_AMOUNT
from
NameOfYourTable
group by
DATE
Order by
DATE
I hope this helps
Each date considers each row for inclusion in the summation of value
SELECT d.DATE, SUM(m.VALUE) AS OUTSTANDING_AMOUNT
FROM yourTable AS d JOIN yourtable AS m ON d.DATE >= m.MATURITY
GROUP BY d.DATE
ORDER BY d.DATE
A possible solution with a tally (numbers) table
SELECT date, SUM(value) outstanding_amount
FROM
(
SELECT date + maturity - n.n date, value, maturity
FROM table1 t JOIN
(
SELECT 1 n UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5
) n ON n.n <= maturity
) q
GROUP BY date
Output:
| DATE | OUTSTANDING_AMOUNT |
-----------------------------
| 1 | 25 |
| 2 | 35 |
| 3 | 15 |
Here is SQLFiddle demo