I want to do other 2 columns that shows me, respectively, the first date and the last day of the week grouped.
Currently, I'm summarizing a query by week period. When I run this query, it shows me the results on table:
SELECT
pc.date,
CONCAT(YEAR(pc.date), '/', WEEK(pc.date)) as year_week
FROM pc
GROUP BY CONCAT(YEAR(pc.date), '/', WEEK(pc.date))
ORDER BY pc.date
date
year_week
2020-09-02
2020/35
2020-09-07
2020/36
2020-09-17
2020/37
2020-09-23
2020/38
2020-09-28
2020/39
2020-10-10
2020/40
2020-10-11
2020/41
2020-10-21
2020/42
2020-10-28
2020/43
How can I find the first and last day of grouped week?
You can use the WEEKDAY function. Demo:
select
date_add(dt, interval -WEEKDAY(dt)-1 day ) FirstDayOfWeek,
date_add(date_add(dt, interval -WEEKDAY(dt)-1 day), interval 6 day) LastDayOfWeek,
week(dt) wk
from (
select '2020-09-02' dt union all
select '2020-09-07' union all
select '2020-09-17'
) t
Returns
FirstDayOfWeek LastDayOfWeek wk
2020-08-30 2020-09-05 35
2020-09-06 2020-09-12 36
2020-09-13 2020-09-19 37
Related
I am trying to write a query to get the last 4 weeks (Mon-Sun) of data. I want every week of data to be stored with an individual and shared table.
every week data store based on name if same name repeated on single week amt should sum and if multiple name it should be show data individual, To see an example of what I am looking for, I have included the desired input and output below.
this is my table
date
amt
name
2022-04-29
5
a
2022-04-28
10
b
2022-04-25
11
a
2022-04-23
15
b
2022-04-21
20
b
2022-04-16
20
a
2022-04-11
10
a
2022-04-10
5
b
2022-04-05
5
b
i want output like this
date
sum(amt)
name
2022-04-25 to 2020-04-29
16
a
2022-04-25 to 2020-04-29
10
b
2022-04-18 to 2022-04-24
35
b
2022-04-11 to 2022-04-17
30
a
2022-04-04 to 2022-04-10
10
b
I would appreciate any pointers or 'best-practises' which I should employ to achieve this task.
You can try to use DATE_ADD with WEEKDAY get week first day and end day.
SELECT
CASE WHEN
weekofyear(`date`) = weekofyear(NOW())
THEN 'current week'
ELSE
CONCAT(date_format(DATE_ADD(`date`, interval - WEEKDAY(`date`) day), '%Y-%m-%d'),' to ',date_format(DATE_ADD(DATE_ADD(`date`, interval -WEEKDAY(`date`) day), interval 6 day), '%Y-%m-%d'))
END 'date',
SUM(amt)
FROM T
GROUP BY
CASE WHEN
weekofyear(`date`) = weekofyear(NOW())
THEN 'current week'
ELSE
CONCAT(date_format(DATE_ADD(`date`, interval - WEEKDAY(`date`) day), '%Y-%m-%d'),' to ',date_format(DATE_ADD(DATE_ADD(`date`, interval -WEEKDAY(`date`) day), interval 6 day), '%Y-%m-%d'))
END
sqlfiddle
EDIT
I saw you edit your question, you can just add name in group by
SELECT
CONCAT(date_format(DATE_ADD(`date`, interval - WEEKDAY(`date`) day), '%Y-%m-%d'),' to ',date_format(DATE_ADD(DATE_ADD(`date`, interval -WEEKDAY(`date`) day), interval 6 day), '%Y-%m-%d')) 'date',
SUM(amt),
name
FROM T
GROUP BY
CONCAT(date_format(DATE_ADD(`date`, interval - WEEKDAY(`date`) day), '%Y-%m-%d'),' to ',date_format(DATE_ADD(DATE_ADD(`date`, interval -WEEKDAY(`date`) day), interval 6 day), '%Y-%m-%d')),
name
ORDER BY 1 desc
sqlfiddle
This is in SQL Server, and just a mess about. Hopefully it can be of some help.
with cteWeekStarts
as
(
select
n,dateadd(week,-n,DATEADD(week, DATEDIFF(week, -1, getdate()), -1)) as START_DATE
from
(values (1),(2),(3),(4)) as t(n)
), cteStartDatesAndEndDates
as
(
select *,dateadd(day,-1,lead(c.start_date) over (order by c.n desc)) as END_DATE
from cteWeekStarts as c
)
,cteSalesSumByDate
as
(
select s.SalesDate,sum(s.salesvalue) as sum_amt from
tblSales as s
group by s.SalesDate
)
select c3.n as WeekNum,c3.START_DATE,isnull(c3.END_DATE,
dateadd(day,6,c3.start_date)) as END_DATE,
(select sum(c2.sum_amt) from cteSalesSumByDate as c2 where c2.SalesDate
between c3.START_DATE and c3.END_DATE) as AMT
from cteStartDatesAndEndDates as c3
order by c3.n desc
I have a table tbl with column date sales
I perform this MySQL query:
SELECT DAY(date) as DAY, count(sales) as sales_count
FROM tbl
I get this result:
DAY sales_count
1 224
2 361
3 155
4 281
.......
30 416
31 332
Now, I find the DAY in WEEK when the sales were MAX by doing this:
SELECT DAY(date) as DAY, WEEK(date) as WEEK, max(count(sales)) as MAX_SALES,
FROM tbl
GROUP BY WEEK(date)
and I get this:
DAY WEEK MAX_SALES
1 1 324
8 2 461
15 3 255
22 4 481
.......
The WEEK and MAX_SALES are correct but the DAY value seems to be the first DAY of the WEEK and not the DAY of the MAX_SALES occurred. I can confirm it is because we have GROUP BY WEEK and that is how DAY is getting the value related to WEEK. We cannot group by MAX()
How do I have the DAY value reflect the MAX_SALES date in the WEEK?
You can try to use subquery.
SELECT weekDay,
DAY,
MAX(sales_count)
FROM (
SELECT DAY(date) as DAY,
Week(date) as weekDay
count(sales) as sales_count
FROM tbl
GROUP BY
DAY(date),
Week(date)
) t1
group by
weekDay,
DAY
I have a database table visitors with three columns:
id | Name | checkin_date |
1 | Reg | 2018-04-20T08:28:54.446Z |
2 | Meg | 2018-04-21T08:28:54.446Z |
3 | Ted | 2018-04-21T08:28:54.446Z |
4 | Bin | 2018-04-23T08:28:54.446Z |
There are several records such as these.
I want to fetch the count of records per each day for only the past 7 days. Right now i was able to fetch the count of visitors per day for all the dates using :
select count(id) as no_of_users
, DATE_FORMAT(checkin_date, '%d %b, %Y') as date
from visitors
GROUP
BY DATE(checkin_date)
But this displays the count of users per each day of all the records. How to get the records of only past 7 days.
select count(id) as no_of_users, DATE_FORMAT(checkin_date, '%d %b, %Y') as date from visitors
where checkin_date >= DATE(NOW()) - INTERVAL 7 DAY
GROUP BY DATE(checkin_date)
in the where is where you want to do the date field >= last 7 days
From your question.
You need to create a calendar table, then LEFT JOIN on the calendar table.
SELECT DATE(t.dt),count(t1.id) cnt
FROM
(
SELECT NOW() dt
UNION ALL
SELECT NOW() - INTERVAL 1 DAY
UNION ALL
SELECT NOW() - INTERVAL 2 DAY
UNION ALL
SELECT NOW() - INTERVAL 3 DAY
UNION ALL
SELECT NOW() - INTERVAL 4 DAY
UNION ALL
SELECT NOW() - INTERVAL 5 DAY
UNION ALL
SELECT NOW() - INTERVAL 6 DAY
UNION ALL
SELECT NOW() - INTERVAL 7 DAY
) t LEFT JOIN T t1
ON DATE(t.dt) = DATE(t1.checkin_date)
group by t1.name,DATE(t.dt)
sqlfiddle:http://sqlfiddle.com/#!9/59f49b/5
select id, count(id) as TOTAL, min (checkin_date) as no_of_users
from visitors
where checkin_date between '<Start Date>' and '<End Date>'
GROUP
BY Id,checkin_date
I'm using this query to calculate the login time of a user on the app for the whole day and previous 5 days
Select
sec_to_time(sum(time_to_sec(TIMEDIFF((IFNULL(logoff_time, ADDTIME(now(), '05:00:00'))),login_time)))) as online_time
from tb_sessions
WHERE
(login_time BETWEEN DATE(DATE_ADD(now(), INTERVAL (-6) DAY))
AND
ADDTIME(now(), '5:00:00')) AND user_id = 30982
AND TIME(`login_time`) between "00:00:00" AND "23:59:59"
group by DATE(login_time)
Now i have some new requirements:
Calculate time from 07:00:00 to 23:59:59
My Table: tb_sessions
id | user_id | login_time | logoff_time
1 3098 2017-06-10 06:30:00 2017-06-10 07:45:00
2 3098 2017-06-10 07:45:01 2017-06-10 08:30:00
By using above query total oline time is = 02:00:00
But i want only time from 7:00 to 8:30, so total time will be = 1:30:00
I make some changes in query with cases but no success.
You can check my query on the below link:
http://sqlfiddle.com/#!9/4620af/12
You could use greatest to take the latest of the dates login_time and 7:00 on the same day, and then use greatest again to exclude negative time differences (when also logoff time is before 7:00):
Select date(login_time) date,
time_format(sec_to_time(sum(greatest(0, time_to_sec(timediff(
ifnull(logoff_time, now()),
greatest(login_time, date_add(date(login_time), interval 7 hour))
))))), '%H:%i:%s') online
from tb_sessions
where login_time between date(date_add(now(), interval (-3) day)) and now()
and user_id = 3098
and time(login_time) between "00:00:00" and "23:59:59"
group by date(login_time)
See it run on sqlfiddle
Explanation
The inner greatest call looks like this:
greatest(login_time, date_add(date(login_time), interval 7 hour))
The second argument takes the date-only from the login_time, so it corresponds to midnight of that day, and then adds 7 hours to it: so this represents 7:00 on that day. greatest will return the latest of these two timestamps. If the first argument represents a time than 7:00, it will be returned. If not, the second argument (i.e. 7:00) will be returned.
The outer greatest call looks like this:
greatest(0, time_to_sec(timediff(....)))
This will make sure the time difference is not negative. Take this example record:
login_time | logoff_time
----------------+----------------
2017-06-01 6:30 | 2017-06-01 6:45
In this case the innermost greatest will return 2017-06-01 7:00, because 6:30 is too early. But that will make timediff() return a negative time interval: -15 minutes. What we really want is 0, because there is no time the user was logged on after 7:00. This is what greatest will do: greatest(0, -15) = 0, so the negative value will be eliminated and will not influence the sum.
Condition on login_time
I left the condition time(login_time) between "00:00:00" and "23:59:59" there, but it really does not do anything, since that is true for all times (unless they are null, but then they would not pass the first condition either).
Edit after New Requirements
In comments you asked how to group by each day when a user doesn't log off on the same day but stays online until 1 or 2 days later.
In that case you need a helper table that will list all days you want to see in the output. This could for instance be seven records for the 7 last days.
Then you have to join your table with it so that there is at least an overlap of the user's session with such a reference date. The calculation of the online time will have to take into account that the log off time might not be before mid night.
Here is the updated query:
select ref_date date,
time_format(sec_to_time(sum(greatest(0, time_to_sec(timediff(
least(ifnull(logoff_time, now()), date_add(ref_date, interval 1 day ), now()),
greatest(login_time, date_add(ref_date, interval 7 hour))
))))), '%H:%i:%s') online
from ( select date(date_add(now(), interval (-6) DAY)) as ref_date union all
select date(date_add(now(), interval (-5) DAY)) union all
select date(date_add(now(), interval (-4) DAY)) union all
select date(date_add(now(), interval (-3) DAY)) union all
select date(date_add(now(), interval (-2) DAY)) union all
select date(date_add(now(), interval (-1) DAY)) union all
select date(now())
) ref
inner join tb_sessions
on login_time < date_add(ref_date, interval 1 day)
and logoff_time > date_add(ref_date, interval 7 hour)
where user_id = 3098
group by ref_date
See it run on sqlfiddle.
I seem to be having a bit of trouble coming up a query to achieve what I want. I have a table like the following..
| Date(TIMESTAMP) | Count |
|---------------------|-------|
| 2016-02-01 01:00:00 | 52 |
| 2016-01-05 11:30:00 | 14 |
| 2016-02-01 04:20:00 | 36 |
| ... | ... |
The table has about 40,000 rows. What I would like to do is grab the totals for multiple date ranges so I end up with the following...
| Period | Total |
|------------|-------|
| All | 10245 |
| Past year | 1401 |
| Past month | 104 |
| Past week | 26 |
Currently I am running through a loop in my PHP script and doing an individual query for each date range I'm looking for. Actually there are about 10 queries I'm doing per loop to grab different stats but for the example I'm simplifying it. This takes forever and I am hoping there is a more elegant way to do this, however I've spent quite a bit of time now trying different things and researching and have gotten nowhere. I understand how to use CASE to group but not when a record may need to be in multiple bins. Any help?
Try this UNION query:
SELECT 'All', COUNT(*) AS Total FROM yourTable
UNION
SELECT 'Past year', COUNT(*) AS Total
FROM yourTable
WHERE DATE(TIMESTAMP) > DATE_ADD(NOW(), INTERVAL -1 YEAR)
UNION
SELECT 'Past month', COUNT(*) AS Total
FROM yourTable
WHERE DATE(TIMESTAMP) > DATE_ADD(NOW(), INTERVAL -1 MONTH)
UNION
SELECT 'Past week', COUNT(*) AS Total
FROM yourTable
WHERE DATE(TIMESTAMP) > DATE_ADD(NOW(), INTERVAL -1 WEEK)
1st. get known to function getting first date of year, first date of month and first date of week.
Then compose your sql using count and filter with first and last date of different period.
ref:
MySQL Select First Day of Year and Month
month
https://stackoverflow.com/a/19259159/1258492
week https://stackoverflow.com/a/11831133/1258492
select 'All' as period, count(1) from
tbl
union
select 'Past Year' as period, count(1) from
tbl
where timestamp between
MAKEDATE(year(now())-1,1) and
last_day(MAKEDATE(year(now())-1,1) + interval 11 month)
union
select 'Past Month' as period, count(1) from
tbl
where timestamp between
LAST_DAY(NOW() - INTERVAL 2 MONTH) + INTERVAL 1 DAY and
LAST_DAY(NOW() - INTERVAL 1 MONTH)
union
select 'Past Week' as period, count(1) from
tbl
where timestamp between
adddate(curdate(), INTERVAL 1-DAYOFWEEK(curdate())-7 DAY) and
adddate(curdate(), INTERVAL 7-DAYOFWEEK(curdate())-7 DAY) ;
You may use subqueries. Use one subquery per time breakdown like so:
SELECT everything, 'past year'
FROM
(
SELECT sum(c) AS 'everything'
FROM reports
) t1,
(
SELECT sum(c) AS 'past year'
FROM reports
WHERE d >= DATE_ADD(CURDATE(), INTERVAL -1 YEAR)
) t2