I am trying to add the total hours where flag =1. Here is how my data look like in the table. The hours are 30 minutes interval
ID FullDateTime Flag
22 2015-02-26 05:30:00.000 1
44 2015-02-26 05:00:00.000 1
25 2015-02-26 04:30:00.000 0
23 2015-02-26 04:00:00.000 1
74 2015-02-26 03:30:00.000 1
36 2015-02-26 03:00:00.000 0
here is what i tried but not working:
select DATEDIFF(minute, sum(FullDatetime), sum(FullDatetime)) / 60.0 as hours
from myTable
where flag = 1
I am expecting the results to be 2 hours.
If the total number of hours is just a function of the number of half hour periods flagged as 1 then a simple count(*) of the rows matching flag 1 multiplied by 0.5 (for the half hour) should do it:
select count(*) * 0.5 from myTable where flag = 1
Related
I have the following need but I am not able to get an effective query:
ID
DATE
PARCEL
STATUS
TYPE
DT_PAY
DT
1
2021-10-15
28
3
R
2021-10-15
2021-10-15
2
2021-11-15
29
0
R
1900-01-01
2021-11-15
3
2021-12-15
30
3
R
2021-12-15
2021-12-15
4
2022-01-15
31
3
R
2022-01-15
2022-01-15
5
2022-02-15
32
3
R
2022-02-15
2022-02-15
6
2022-03-15
33
0
R
1900-01-01
2022-03-15
7
2022-04-15
34
0
R
1900-01-01
2022-04-15
8
2022-05-15
35
0
R
1900-01-01
2022-05-15
9
2022-06-15
36
0
R
1900-01-01
2022-06-15
10
2022-07-15
37
3
R
2022-07-15
2022-07-15
With the data in the table above you would need the following result:
ID
DATE
PARCEL
STATUS
TYPE
DT_PAY
DT
6
2022-03-15
33
0
R
1900-01-01
2022-03-15
2
2021-11-15
29
0
R
1900-01-01
2021-11-15
It is necessary to list the first occurrence of a line where STATUS = 0 appears after a line with STATUS = 3 appears, and the second time this occurs after another line appears with STATUS = 3 as well, but being from the most current to the oldest date, in this case the date 2022-03-15 is more current and the date 2021-11-15 is more old one that meets the STATUS = 0 filter appears after a line with STATUS = 3 appears
My query only works to find STATUS=3, but needed it to be the same for STATUS=0
with TopDates as
(select row_number() over (order by DT desc) as Row, *
from DBO.TABLE
WHERE DT < GETDATE ()
AND DT_PAY <> '1900-01-01'
AND STATUS = '3'
)
select
TB.ID
,TB.DATE
,TB.PARCEL
,TB.STATUS
,TB.DT_PAY
,TB.DT
from TopDates TB
where Row<=2
Just add an OR clause in there?
Or am I not understanding you correctly?
`with TopDates as
(select row_number() over (order by DT desc) as Row, *
from DBO.TABLE
WHERE DT < GETDATE ()
AND DT_PAY <> '1900-01-01'
AND STATUS = '3'
OR STATUS = '0'
)
select
TB.ID
,TB.DATE
,TB.PARCEL
,TB.STATUS
,TB.DT_PAY
,TB.DT
from TopDates TB
where Row<=2
So my data looks like this:
DATE TEMPERATURE
2012-01-13 23:15:00 UTC 0
2012-01-14 01:35:00 UTC 5
2012-01-14 02:15:00 UTC 6
2012-01-14 03:15:00 UTC 8
2012-01-14 04:15:00 UTC 0
2012-01-14 04:55:00 UTC 0
2012-01-14 05:15:00 UTC -2
2012-01-14 05:35:00 UTC 0
I am trying to calculate the amount of time a zip code temperature will drop to 0 or below on any given day. On the 13th, it only happens for a very short amount of time so we don't really care. I want to know how to calculate the number of minutes this happens on the 14th, since it looks like a significantly (and consistently) cold day.
I want the query to add two more columns.
The first column added would be the time difference between the rows on a given date. So row 3- row 2=40 mins and row 4-row3=60 mins.
The second column would total the amount of minutes for a whole day the minutes the temperature has dropped to 0 or below. Here row 2-4 would be ignored. From row 5-8, total time that the temperature was 0 or below would be about 90 mins
It should end up looking like this:
DATE TEMPERATURE MINUTES_DIFFERENCE TOTAL_MINUTES
2012-01-13 23:15:00 UTC 0 0 0
2012-01-14 01:35:00 UTC 5 140 0
2012-01-14 02:15:00 UTC 6 40 0
2012-01-14 03:15:00 UTC 8 60 0
2012-01-14 04:15:00 UTC 0 60 60
2012-01-14 04:55:00 UTC 0 30 90
2012-01-14 05:15:00 UTC-2 20 110
2012-01-14 05:35:00 UTC 0 20 130
Use below
select *,
sum(minutes_difference) over(order by date) total_minutes
from (
select *,
ifnull(timestamp_diff(timestamp(date), lag(timestamp(date)) over(order by date), minute), 0) as minutes_difference
from your_table
)
if applied to sample data in your question - output is
Update to answer updated question
select * except(new_grp, grp),
sum(if(temperature > 0, 0, minutes_difference)) over(partition by grp order by date) total_minutes
from (
select *, countif(new_grp) over(order by date) as grp
from (
select *,
ifnull(timestamp_diff(timestamp(date), lag(timestamp(date)) over(order by date), minute), 0) as minutes_difference,
ifnull(((temperature <= 0) and (lag(temperature) over(order by date) > 0)) or
((temperature > 0) and (lag(temperature) over(order by date) <= 0)), true) as new_grp
from your_table
)
)
with output
I have data like following
ID SalesTime Qty Unit Price Item
1 01/01/2021 08:10:00 10 10 A
2 01/01/2021 11:30:00 2 9 B
3 01/01/2021 11:59:50 1 8 C
4 01/02/2021 13:00:00 5 15 D
5 01/03/2021 10:00:00 4 10 A
6 01/03/2021 12:00:00 5 9 B
7 01/03/2021 12:50:00 6 15 D
8 01/04/2021 10:50:00 5 8 C
9 01/04/2021 11:10:00 2 10 A
10 ............
I wanna summarize the total into the form,
for example:
Mon Tue Wed Thu Fri Sat Sun
08:00~09:59 20 21 50 100 60 70 210
10:00~11:59 60 25 60 90 75 80 200
12:00~13:59 100 10 50 60 70 50 150
How to do that in MS SQL, thanks a lot.
You can extract the hour and divide by two for the rows. And then use conditional aggregation for the columns. Assuming you want the total of the price times quantity:
select convert(time, dateadd(hour, 2 * (datepart(hour, salestime) / 2), 0)) as hh,
sum(case when datename(weekday, salestime) = 'Monday' then qty * unit_price end) as mon,
sum(case when datename(weekday, salestime) = 'Tuesday' then qty * unit_price end) as tue,
. . .
from t
group by datepart(hour, salestime) / 2
order by min(salestime);
Note: This just returns the beginning of the time period, rather than the full range.
I have a running balance sheet showing customer balances after inflows and (outflows) by date. It looks something like this:
ID DATE AMOUNT RUNNING AMOUNT
-- ---------------- ------- --------------
10 27/06/2019 14:30 100 100
10 29/06/2019 15:26 -100 0
10 03/07/2019 01:56 83 83
10 04/07/2019 17:53 15 98
10 05/07/2019 15:09 -98 0
10 05/07/2019 15:53 98.98 98.98
10 05/07/2019 19:54 -98.98 0
10 07/07/2019 01:36 90.97 90.97
10 07/07/2019 13:02 -90.97 0
10 07/07/2019 16:32 39.88 39.88
10 08/07/2019 13:41 50 89.88
20 08/01/2019 09:03 890.97 890.97
20 09/01/2019 14:47 -91.09 799.88
20 09/01/2019 14:53 100 899.88
20 09/01/2019 14:59 -399 500.88
20 09/01/2019 18:24 311 811.88
20 09/01/2019 23:25 50 861.88
20 10/01/2019 16:18 -861.88 0
20 12/01/2019 16:46 894.49 894.49
20 25/01/2019 05:40 -871.05 23.44
I have attempted using lag() but I seem not to understand how to use it yet.
SELECT ID, MEDIAN(DIFF) MEDIAN_AGE
FROM
(
SELECT *, DATEDIFF(day, Lag(DATE, 1) OVER(ORDER BY ID), DATE
)AS DIFF
FROM TABLE 1
WHERE RUNNING AMOUNT = 0
)
GROUP BY ID;
The expected result would be:
ID MEDIAN_AGE
-- ----------
10 1
20 2
Please help in writing out the query that gives the expected result.
As already pointed out, you are using syntax that isn't valid for Oracle, including functions that don't exist and column names that aren't allowed.
You seem to want to calculate the number of days between a zero running-amount and the following non-zero running-amount; lead() is probably easier than lag() here, and you can use a case expression to only calculate it when needed:
select id, date_, amount, running_amount,
case when running_amount = 0 then
lead(date_) over (partition by id order by date_) - date_
end as diff
from your_table;
ID DATE_ AMOUNT RUNNING_AMOUNT DIFF
---------- -------------------- ---------- -------------- ----------
10 2019-06-27 14:30:00 100 100
10 2019-06-29 15:26:00 -100 0 3.4375
10 2019-07-03 01:56:00 83 83
10 2019-07-04 17:53:00 15 98
10 2019-07-05 15:09:00 -98 0 .0305555556
10 2019-07-05 15:53:00 98.98 98.98
10 2019-07-05 19:54:00 -98.98 0 1.2375
10 2019-07-07 01:36:00 90.97 90.97
10 2019-07-07 13:02:00 -90.97 0 .145833333
10 2019-07-07 16:32:00 39.88 39.88
10 2019-07-08 13:41:00 50 89.88
20 2019-01-08 09:03:00 890.97 890.97
20 2019-01-09 14:47:00 -91.09 799.88
20 2019-01-09 14:53:00 100 899.88
20 2019-01-09 14:59:00 -399 500.88
20 2019-01-09 18:24:00 311 811.88
20 2019-01-09 23:25:00 50 861.88
20 2019-01-10 16:18:00 -861.88 0 2.01944444
20 2019-01-12 16:46:00 894.49 894.49
20 2019-01-25 05:40:00 -871.05 23.44
Then use the median() function, rounding if desired to get your expected result:
select id, median(diff) as median_age, round(median(diff)) as median_age_rounded
from (
select id, date_, amount, running_amount,
case when running_amount = 0 then
lead(date_) over (partition by id order by date_) - date_
end as diff
from your_table
)
group by id;
ID MEDIAN_AGE MEDIAN_AGE_ROUNDED
---------- ---------- ------------------
10 .691666667 1
20 2.01944444 2
db<>fiddle
How to find the date difference in hours between two records with nearest datetime value and it must be compared in same group?
Sample Data as follows:
Select * from tblGroup
Group FinishedDatetime
1 03-01-2009 00:00
1 13-01-2009 22:00
1 08-01-2009 03:00
2 01-01-2009 10:00
2 13-01-2009 20:00
2 10:01-2009 10:00
3 27-10-2008 00:00
3 29-10-2008 00:00
Expected Output :
Group FinishedDatetime Hours
1 03-01-2009 00:00 123
1 13-01-2009 22:00 139
1 08-01-2009 03:00 117
2 01-01-2009 10:00 216
2 13-01-2009 20:00 82
2 10:01-2009 10:00 82
3 27-10-2008 00:00 48
3 29-10-2008 00:00 48
Try this:
Select t1.[Group], DATEDIFF(HOUR, z.FinishedDatetime, t1.FinishedDatetime)
FROM tblGroup t1
OUTER APPLY(SELECT TOP 1 *
FROM tblGroup t2
WHERE t2.[Group] = t1.[Group] AND t2.FinishedDatetime<t1.FinishedDatetime
ORDER BY FinishedDatetime DESC)z