I'm trying to calculate the total working hours between two dates in bigquery sql:
The dates being between MAX(date) and DATE_SUB(MAX(date), interval 3 month).
In other words, I want to know the sum of working hours between the exit date and 3 months prior to the exit date.
The current table is something like this:
id
date
hours
abc
2020-10-01
12
abc
2020-12-07
4
abc
2020-12-12
12
abc
2020-12-25
6
abc
2021-01-07
9
abc
2021-02-04
7
The ideal output is:
id
hours
abc
38
I have multiple workers and workers have different working dates and hours.
We need a subquery here to calculate exit_date first:
with mytable as (
select 'abc' as id, DATE '2020-10-01' as date, 12 as hours union all
select 'abc' as id, DATE '2020-12-07' as date, 4 as hours union all
select 'abc' as id, DATE '2020-12-12' as date, 12 as hours union all
select 'abc' as id, DATE '2020-12-25' as date, 6 as hours union all
select 'abc' as id, DATE '2021-01-07' as date, 9 as hours union all
select 'abc' as id, DATE '2021-02-04' as date, 7 as hours
)
select
id,
sum(hours) as hours
from (
select *, MAX(date) OVER (PARTITION BY id) as exit_date
from mytable
)
where date >= DATE_SUB(exit_date, INTERVAL 3 MONTH)
group by id
Assume we have exampleDB and select contractdate like
SELECT DB.contractdate
FROM exampleDB DB
contractdate
2014/12/1
2015/12/1
2016/12/1
2017/12/1
2018/12/1
2019/12/1
I would like to count the policy number at each time like
each time policy count
2014/1/1 0
2015/1/1 1
2016/1/1 2
2017/1/1 3
2018/1/1 4
2019/1/1 5
I tried
WHERE DB.contractdate <='2014/1/1';
But I must loop such code manually.
How can I loop?
If the binning is every month,it is very stressful process.
can they be combined into one?
Best regards
select contractdate as "each time",
count(*) as "policy count"
from exampleDB
where contractdate in (mention all dates you want)
group by contractdate
Hope this will help you.
you can use row_num() and trunc() to get 1st day of the month
SELECT TRUNC(DB.contractdate, 'MONTH'), row_number() over (order by DB.contractdate) -1 as policy_count
FROM exampleDB DB
You can use COUNT analytical function with RANGE operator as follows:
SQL> with dataa(contractdate) as
2 (
3 select date '2014-12-01' from dual union all
4 select date '2015-12-01' from dual union all
5 select date '2016-12-01' from dual union all
6 select date '2017-12-01' from dual union all
7 select date '2018-12-01' from dual union all
8 select date '2019-12-01' from dual
9 )
10 SELECT
11 TRUNC(CONTRACTDATE, 'year') as "each time",
12 COUNT(1) OVER(
13 ORDER BY
14 CONTRACTDATE DESC
15 RANGE BETWEEN 1 FOLLOWING AND UNBOUNDED FOLLOWING
16 ) as "policy count"
17 FROM
18 DATAA order by 1;
each time policy count
--------- ------------
01-JAN-14 0
01-JAN-15 1
01-JAN-16 2
01-JAN-17 3
01-JAN-18 4
01-JAN-19 5
6 rows selected.
SQL>
Cheers!!
I want to make a query, which shows the progress of the number of users on my webpage by week.
I use following query to run the users database and get the number, grouped by a week:
SELECT TRUNC(FAB.LICENSE_DATE, 'IW'),
COUNT(DISTINCT FAB.STATEMENT_NUMBER) AS "Number of account statements"
FROM USERS FAB
GROUP BY TRUNC(FAB.LAST_UPDATED_TIME, 'IW');
This gives following output:
Date | Users
----------------------
2015/09/07 | 5
2015/09/14 | 4
2015/09/21 | 6
But this is actually not what I want to achieve. I want to have the following output:
Date | Users
----------------------
2015/09/07 | 5
2015/09/14 | 9 (5 + 4)
2015/09/21 | 15 (5 + 4 + 6)
How to modify the query so I get all the results?
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE USERS (
LICENSE_DATE,
LAST_UPDATED_TIME,
STATEMENT_NUMBER
) AS
SELECT DATE '2015-09-07', DATE '2015-09-07', 1 FROM DUAL
UNION ALL SELECT DATE '2015-09-08', DATE '2015-09-08', 2 FROM DUAL
UNION ALL SELECT DATE '2015-09-08', DATE '2015-09-08', 3 FROM DUAL
UNION ALL SELECT DATE '2015-09-09', DATE '2015-09-09', 4 FROM DUAL
UNION ALL SELECT DATE '2015-09-12', DATE '2015-09-12', 5 FROM DUAL
UNION ALL SELECT DATE '2015-09-14', DATE '2015-09-15', 6 FROM DUAL
UNION ALL SELECT DATE '2015-09-15', DATE '2015-09-16', 7 FROM DUAL
UNION ALL SELECT DATE '2015-09-16', DATE '2015-09-16', 8 FROM DUAL
UNION ALL SELECT DATE '2015-09-17', DATE '2015-09-18', 9 FROM DUAL
UNION ALL SELECT DATE '2015-09-21', DATE '2015-09-21', 10 FROM DUAL
UNION ALL SELECT DATE '2015-09-21', DATE '2015-09-26', 11 FROM DUAL
UNION ALL SELECT DATE '2015-09-22', DATE '2015-09-22', 12 FROM DUAL
UNION ALL SELECT DATE '2015-09-23', DATE '2015-09-25', 13 FROM DUAL
UNION ALL SELECT DATE '2015-09-24', DATE '2015-09-24', 14 FROM DUAL
UNION ALL SELECT DATE '2015-09-27', DATE '2015-09-27', 15 FROM DUAL;
Query 1:
SELECT LAST_UPDATED_WEEK,
SUM( NUM_STATEMENTS ) OVER ( ORDER BY LAST_UPDATED_WEEK ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW ) AS "Number of account statements"
FROM (
SELECT TRUNC(LAST_UPDATED_TIME, 'IW') AS LAST_UPDATED_WEEK,
COUNT(DISTINCT STATEMENT_NUMBER) AS NUM_STATEMENTS
FROM USERS
GROUP BY
TRUNC( LAST_UPDATED_TIME, 'IW')
)
Results:
| LAST_UPDATED_WEEK | Number of account statements |
|-----------------------------|------------------------------|
| September, 07 2015 00:00:00 | 5 |
| September, 14 2015 00:00:00 | 9 |
| September, 21 2015 00:00:00 | 15 |
SELECT TRUNC(FAB.LICENSE_DATE, 'IW'),
SUM(COUNT(DISTINCT FAB.STATEMENT_NUMBER)) OVER (ORDER BY TRUNC(FAB.LAST_UPDATED_TIME, 'IW')) as "Number of account statements"
FROM USERS FAB
GROUP BY TRUNC(FAB.LAST_UPDATED_TIME, 'IW');
You can use this code block for your problem :
select u.date
,(select sum(u1.users)
from users u1
where u1.ddate <= u.date) as users
from users u;
It gives this output :
07.09.2015 5
14.09.2015 9
21.09.2015 15
Good luck
Hello you can try this code too.
WITH t1 AS
( SELECT to_date('01/01/2015','mm/dd/yyyy') rn, 5 usrs FROM dual
UNION ALL
SELECT to_date('02/01/2015','mm/dd/yyyy') rn, 4 usrs FROM dual
UNION ALL
SELECT to_date('03/01/2015','mm/dd/yyyy') rn, 8 usrs FROM dual
UNION ALL
SELECT to_date('04/01/2015','mm/dd/yyyy') rn, 2 usrs FROM dual
)
SELECT rn,
usrs,
sum(usrs) over (order by rn ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) cumm_usrs
FROM t1
GROUP BY rn,
usrs;
This is more of a best practices question although I am struggling with the sql but the 2 are linked. I want to insert into a sql database the date and time for the forthcoming 24 hour period starting at 6am.
I think it would be best to run a schedualed SQL job at say 1am for the forthcoming day. This would create one column and 24 rows spanning for example 20/03/2013 06:00 to 21/03/2013 05:00.
thanks
The problem is the date time functions. Here is an example of how it could be done (in SQL Server):
insert into t(thedatetime)
select dateadd(hour, hrs.hr, cast(CAST(getdate() as DATE) as datetime))
from (select 0 as hr union all select 1 union all select 2 union all select 3 union all
select 4 union all select 5 union all select 6 union all select 7 union all
select 8 union all select 9 union all select 10 union all select 11 union all
select 12 union all select 13 union all select 14 union all select 15 union all
select 16 union all select 17 union all select 18 union all select 19 union all
select 20 union all select 21 union all select 22 union all select 23
) as hrs;
In Oracle, the select would might be:
select trunc(sysdate) + hrs.hr/24.0
And there are similar constructs for other databases.
This is assuming that it is running after midnight on the date in question. For the next day, you would have to add one day to the current date.
This works in Postgres:
select timestamp '2013-03-30 06:00' + interval '1' hour * i
from generate_series(0,23) i;
This will work for ORACLE
SELECT To_date('22-03-2013 '||(CASE WHEN (6+(LEVEL-1))>12 THEN (CASE WHEN LEVEL>19 THEN (LEVEL-12-7) ELSE (6+(LEVEL-1))-12 END) ELSE (6+(LEVEL-1)) END)||':00:00','DD-MM-YYYY HH:MI:SS')
FROM dual CONNECT BY LEVEL<=24;
I have the following data being returned from a query. Essentially I am putting this in a temp table so it is now in a temp table that I can query off of(Obviously a lot more data in real life, I am just showing an example):
EmpId Date
1 2011-01-01
1 2011-01-02
1 2011-01-03
2 2011-02-03
3 2011-03-01
4 2011-03-02
5 2011-01-02
I need to return only EmpId's that have 30 or more consecutive days in the date column. I also need to return the day count for these employees that have 30 or more consecutive days. There could potentially be 2 or more sets of different consecutive days that are 30 or more days. iIn this instance I would like to return multiple rows. So if an employee has a date from 2011-01-01 to 2011-02-20 then return this and the count in one row. Then if this same employee has dates of 2011-05-01 to 2011-07-01 then return this in another row. Essentially all breaks in consecutive days are treated as a seperate record.
Using DENSE_RANK should do the trick:
;WITH sampledata
AS (SELECT 1 AS id, DATEADD(day, -0, GETDATE())AS somedate
UNION ALL SELECT 1, DATEADD(day, -1, GETDATE())
UNION ALL SELECT 1, DATEADD(day, -2, GETDATE())
UNION ALL SELECT 1, DATEADD(day, -3, GETDATE())
UNION ALL SELECT 1, DATEADD(day, -4, GETDATE())
UNION ALL SELECT 1, DATEADD(day, -5, GETDATE())
UNION ALL SELECT 1, DATEADD(day, -10, GETDATE())
UNION ALL SELECT 1, '2011-01-01 00:00:00'
UNION ALL SELECT 1, '2010-12-31 00:00:00'
UNION ALL SELECT 1, '2011-02-01 00:00:00'
UNION ALL SELECT 1, DATEADD(day, -10, GETDATE())
UNION ALL SELECT 2, DATEADD(day, 0, GETDATE())
UNION ALL SELECT 2, DATEADD(day, -1, GETDATE())
UNION ALL SELECT 2, DATEADD(day, -2, GETDATE())
UNION ALL SELECT 2, DATEADD(day, -6, GETDATE())
UNION ALL SELECT 3, DATEADD(day, 0, GETDATE())
UNION ALL SELECT 4, DATEADD(day, 0, GETDATE())
UNION ALL SELECT 5, DATEADD(day, 0, GETDATE()))
, ranking
AS (SELECT *, DENSE_RANK()OVER(PARTITION BY id ORDER BY DATEDIFF(day, 0, somedate)) - DATEDIFF(day, 0, somedate)AS dategroup
FROM sampledata)
SELECT id
, MIN(somedate)AS range_start
, MAX(somedate)AS range_end
, DATEDIFF(day, MIN(somedate), MAX(somedate)) + 1 AS consecutive_days
FROM ranking
GROUP BY id, dategroup
--HAVING DATEDIFF(day, MIN(somedate), MAX(somedate)) + 1 >= 30 --change as needed
ORDER BY id, range_start
Something like this should do the trick, haven't tested it though.
SELECT
a.empid
, count(*) as consecutive_count
, min(a.mydate) as startdate
FROM (SELECT * FROM logins ORDER BY mydate) a
INNER JOIN (SELECT * FROM logins ORDER BY mydate) b
ON (a.empid = b.empid AND datediff(day,a.mydate,b.mydate) = 1
GROUP BY a.empid, startdate
HAVING consecutive_count > 30
This is a good case for a recursive CTE. I stole the data table from #Davin:
with data AS --sample data
( SELECT 1 as id ,DATEADD(DD,-0,GETDATE()) as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-1,GETDATE()) as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-2,GETDATE()) as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-3,GETDATE()) as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-4,GETDATE()) as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-5,GETDATE()) as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-10,GETDATE()) as date UNION ALL
SELECT 1 as id ,'2011-01-01 00:00:00.000' as date UNION ALL
SELECT 1 as id ,'2010-12-31 00:00:00.000' as date UNION ALL
SELECT 1 as id ,'2011-02-01 00:00:00.000' as date UNION ALL
SELECT 1 as id ,DATEADD(DD,-10,GETDATE()) as date UNION ALL
SELECT 2 as id ,DATEADD(DD,0,GETDATE()) as date UNION ALL
SELECT 2 as id ,DATEADD(DD,-1,GETDATE()) as date UNION ALL
SELECT 2 as id ,DATEADD(DD,-2,GETDATE()) as date UNION ALL
SELECT 2 as id ,DATEADD(DD,-6,GETDATE()) as date UNION ALL
SELECT 3 as id ,DATEADD(DD,0,GETDATE()) as date UNION ALL
SELECT 4 as id ,DATEADD(DD,0,GETDATE()) as date UNION ALL
SELECT 5 as id ,DATEADD(DD,0,GETDATE()) as date )
,CTE AS
(
SELECT id, CAST(date as date) Date, Consec = 1
FROM data
UNION ALL
SELECT t.id, CAST(t.date as DATE) Date, Consec = (c.Consec + 1)
FROM data T
INNER JOIN CTE c
ON T.id = c.id
AND CAST(t.date as date) = CAST(DATEADD(day, 1, c.date) as date)
)
SELECT id, MAX(consec)
FROM CTE
GROUP BY id
ORDER BY id
Basically this generates a lot of rows per person, and measures how many days in a row each date represents.
Assuming there are no duplicate dates for the same employee:
;WITH ranged AS (
SELECT
EmpId,
Date,
RangeId = DATEDIFF(DAY, 0, Date)
- ROW_NUMBER() OVER (PARTITION BY EmpId ORDER BY Date)
FROM atable
)
SELECT
EmpId,
StartDate = MIN(Date),
EndDate = MAX(Date),
DayCount = DATEDIFF(DAY, MIN(Date), MAX(Date)) + 1
FROM ranged
GROUP BY EmpId, RangeId
HAVING DATEDIFF(DAY, MIN(Date), MAX(Date)) + 1 >= 30
ORDER BY EmpId, MIN(Date)
DATEDIFF turns the dates into integers (the difference of days between the 0 date (1900-01-01) and Date). If the dates are consecutive, the integers are consecutive too. Using the data sample in the question as an example, the DATEDIFF results will be:
EmpId Date DATEDIFF
----- ---------- --------
1 2011-01-01 40542
1 2011-01-02 40543
1 2011-01-03 40544
2 2011-02-03 40575
3 2011-03-01 40601
4 2011-03-02 40602
5 2011-01-02 40543
Now, if you take each employee's rows, assign row numbers to them in the order of dates, and get the difference between the numeric representations and row numbers, you will find that the difference stays the same for consecutive numbers (and, therefore, consecutive dates). Using a slightly different sample for better illustration, it will look like this:
Date DATEDIFF RowNum RangeId
---------- -------- ------ -------
2011-01-01 40542 1 40541
2011-01-02 40543 2 40541
2011-01-03 40544 3 40541
2011-01-05 40546 4 40542
2011-01-07 40548 5 40543
2011-01-08 40549 6 40543
2011-01-09 40550 7 40543
The specific value of RangeId is not important, only the fact that it remains the same for consecutive dates matters. Based on that fact, you can use it as a grouping criterion to count the dates in the group and get the range bounds.
The above query uses DATEDIFF(DAY, MIN(Date), MAX(Date)) + 1 to count the days, but you could also simply use COUNT(*) instead.