For Example,
I had data something like this :-
batch MIN MAX TIME
X 10 20 2018-07-12 10:29:00.000
X 30 50 2018-07-12 10:30:00.000
X 50 30 2018-07-12 10:31:00.000
| | | |
X 40 20 2018-07-12 11:45:00.000
Now I want hourly data based on start time, For example :-
DURATION MIN
2018-07-12 10:29:00.000-2018-07-12 11:29:00.000 10
2018-07-12 11:30:00.000-2018-07-12 12:30:00.000 10
How can I get this?(Get Min Value For every hour based on Start Time)
dateadd function allows you to add or subtract days,hours, minutes to a date.
Consider The below query
select dateadd(HOUR, -1, getdate()) as time_added,
getdate() as curr_date
The -1 is for subtracting one hour (adding negative one hour)
The result of above query is :
timeadded curr_date
2018-07-12 13:25:31.603 2018-07-12 14:25:31.603
Instead of getdate() use your startdate
In your case it would be
select min from table where time<#starttime and time> dateadd(HOUR, -1, #starttime)
Related
I need some help regarding sum of production count for overnight shifts.
The table just contains a timestamp (that is automaticaly generated by SQL server during INSERT), the number of OK produced pieces and the number of NOT OK produced pieces in that given timestamp.
CREATE TABLE [machine1](
[timestamp] [datetime] NOT NULL,
[OK] [int] NOT NULL,
[NOK] [int] NOT NULL
)
ALTER TABLE [machine1] ADD DEFAULT (getdate()) FOR [timestamp]
The table holds values like these (just an example, there are hundreds of lines each day and the time stamps are not fixed like each hour or each 30mins):
timestamp
OK
NOK
2022-08-01 05:30:00.000
15
1
2022-08-01 06:30:00.000
18
3
...
...
...
2022-08-01 21:30:00.000
10
12
2022-08-01 22:30:00.000
0
3
...
...
...
2022-08-01 23:59:00.000
1
2
2022-08-02 00:01:00.000
7
0
...
...
...
2022-08-02 05:30:00.000
12
4
2022-08-02 06:30:00.000
9
3
The production works in shifts like so:
morning shift: 6:00 -> 14:00
afternoon shift: 14:00 -> 22:00
night shift: 22:00 -> 6:00 the next day
I have managed to get sums for the morning and afternoon shifts without issues but I can't figure out how to do the sum for the night shift (I have these SELECTs for each shift stored as a VIEW for easy access).
For the morning shift:
SELECT CAST(timestamp AS date) AS Morning,
SUM(OK) AS SUM_OK,
SUM(NOK) AS SUM_NOK
FROM [machine1]
WHERE DATEPART(hh,timestamp) >= 6 AND DATEPART(hh,timestamp) < 14
GROUP BY CAST(timestamp AS date)
ORDER BY Morning ASC
For the afternoon shift:
SELECT CAST(timestamp AS date) AS Afternoon,
SUM(OK) AS SUM_OK,
SUM(NOK) AS SUM_NOK
FROM [machine1]
WHERE DATEPART(hh,timestamp) >= 14 AND DATEPART(hh,timestamp) < 22
GROUP BY CAST(timestamp AS date)
ORDER BY Afternoon ASC
Since we identify the date of each shift by its start, my idea would be that the result for such SUM of night shift would be
Night
SUM_OK
SUM_NOK
2022-08-01
xxx
xxx
for interval 2022-08-01 22:00:00.000 -> 2022-08-02 05:59:59.999
2022-08-02
xxx
xxx
for interval 2022-08-02 22:00:00.000 -> 2022-08-03 05:59:59.999
2022-08-03
xxx
xxx
for interval 2022-08-03 22:00:00.000 -> 2022-08-04 05:59:59.999
2022-08-04
xxx
xxx
for interval 2022-08-04 22:00:00.000 -> 2022-08-05 05:59:59.999
...
...
...
After few days of trial and error I have probably managed to find the needed solution. Using a subquery I shift all the times in range 00:00:00 -> 05:59:59 to the previous day and then I use that result in same approach as for morning and afternon shift (because now all the production data from night shift are in the same date between 22:00:00 and 23:59:59).
In case anyone needs it in future:
SELECT
CAST(nightShift.shiftedTime AS date) AS Night,
SUM(nightShift.OK) AS SUM_OK,
SUM(nightShift.NOK) AS SUM_NOK
FROM
(SELECT
CASE WHEN (DATEPART(hh, timestamp) < 6 AND DATEPART(hh, timestamp) >= 4) THEN DATEADD(HOUR, -6, timestamp)
WHEN (DATEPART(hh, timestamp) < 4 AND DATEPART(hh, timestamp) >= 2) THEN DATEADD(HOUR, -4, timestamp)
WHEN (DATEPART(hh, timestamp) < 2 AND DATEPART(hh, timestamp) >= 0) THEN DATEADD(HOUR, -2, timestamp)
END AS shiftedTime,
[OK],
[NOK]
FROM [machine1]
WHERE (DATEPART(hh, cas) >= 0 AND DATEPART(hh, cas) < 6)) nightShift
WHERE DATEPART(hh,nightShift.shiftedTime) >= 22
GROUP BY CAST(nightShift.shiftedTime AS date)
ORDER BY Night ASC
PS: If there is anything wrong with this approach, please feel free to correct me as I'm just newbie in SQL. So far this seems to do exactly what I needed.
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 a table with a datetime field. I'm trying to round up/down the time in 15 minute intervals. But with non standard mathematical rounding rules where rounding up happens if its greater than 5 minutes past the 15 minute interval.
For example
IF 06:05 round down to 06:00
IF 06:06 round up to 06:15
IF 06:20 round down to 06:15
IF 06:21 round up to 06:30
and so on..
I've managed to find here T-SQL: Round to nearest 15 minute interval to round the nearest 15 minutes but this uses mathematical rounding meaning 06:07 would still round down to 06:00 instead of rounding up to 06:15.
Below code where i've got to:
cast(datepart(hour, getdate()) + datepart(minute, getdate()) / 60.00 as decimal(5, 2))
Just use a couple of date tricks.
This code will give you the top of the hour for the time you're evaluating (minutes effectively removed by adding up the hours since the 0 date in SQL):
select dateadd(hour, datediff(hour, 0, getdate()), 0)
From there, you need a CASE expression to evaluate which quartile of the hour the time in question falls into (just a snippet here):
case
when datepart(minute, dtm) > 50 then 60
when datepart(minute, dtm) > 35 then 45
when datepart(minute, dtm) > 20 then 30
when datepart(minute, dtm) > 5 then 15
else 0
end
Put those two pieces together with a DATEADD to decide how many minutes we're adding to that even hour mark:
declare #dtms table (dtm datetime);
insert #dtms (dtm)
values ('2019-07-16T12:05:00'),
('2019-07-16T12:06:00'),
('2019-07-16T12:21:00'),
('2019-07-16T12:29:00'),
('2019-07-16T12:35:00'),
('2019-07-16T12:38:00'),
('2019-07-16T12:56:00')
select
dtm,
dateadd(minute,
case
when datepart(minute, dtm) > 50 then 60
when datepart(minute, dtm) > 35 then 45
when datepart(minute, dtm) > 20 then 30
when datepart(minute, dtm) > 5 then 15
else 0
end, dateadd(hour, datediff(hour, 0, dtm), 0)) as rounded
from #dtms;
Results:
+-------------------------+-------------------------+
| dtm | rounded |
+-------------------------+-------------------------+
| 2019-07-16 12:05:00.000 | 2019-07-16 12:00:00.000 |
| 2019-07-16 12:06:00.000 | 2019-07-16 12:15:00.000 |
| 2019-07-16 12:21:00.000 | 2019-07-16 12:30:00.000 |
| 2019-07-16 12:29:00.000 | 2019-07-16 12:30:00.000 |
| 2019-07-16 12:35:00.000 | 2019-07-16 12:30:00.000 |
| 2019-07-16 12:38:00.000 | 2019-07-16 12:45:00.000 |
| 2019-07-16 12:56:00.000 | 2019-07-16 13:00:00.000 |
+-------------------------+-------------------------+
I have a table with daily dates starting from 31st December 1999 up to 31st December 2050, excluding weekends.
Say given a particular date, for this example lets use 2019-03-14. I want to pick the date that was 30 days previous (the number of days needs to be flexible as it won't always be 30), ignoring weekends which in this case would be 2019-02-01.
How to do this?
I wrote the query below & it indeed lists 30 days previous to the specified date.
select top 30 Date
from DateDimension
where IsWeekend = 0 and Date <= '2019-03-14'
order by Date desc
So I thought I could use the query below to get the correct answer of 2019-02-01
;with ds as
(
select top 30 Date
from DateDimension
where IsWeekend = 0 and Date <= '2019-03-14'
)
select min(Date) from ds
However this doesn't work. It returns me the first date in my table, 1999-12-31.
2019-03-14
2019-03-13
2019-03-12
2019-03-11
2019-03-08
2019-03-07
2019-03-06
2019-03-05
2019-03-04
2019-03-01
2019-02-28
2019-02-27
2019-02-26
2019-02-25
2019-02-22
2019-02-21
2019-02-20
2019-02-19
2019-02-18
2019-02-15
2019-02-14
2019-02-13
2019-02-12
2019-02-11
2019-02-08
2019-02-07
2019-02-06
2019-02-05
2019-02-04
2019-02-01
TOP is meaningless without an ORDER BY, so you could do something like
;with ds as
(
select top 30 Date
from DateDimension
where IsWeekend = 0 and Date <= '2019-03-14'
order by Date DESC
)
select min(Date) from ds;
even better would be to use the ANSI syntax instead of TOP:
select Date
from DateDimension
where IsWeekend = 0 and Date <= '2019-03-14'
order by Date DESC
OFFSET 30 ROWS FETCH NEXT 1 ROW ONLY;
DISCLAIMER - code not tested since you did not provide DDL and sample data
HTH
Sql Fiddle Example
I have this result table
Id Hours
----- -----
1 09:00
2 09:30
3 10:00
4 10:30
5 11:00
6 11:30
7 12:00
8 12:30
9 13:00
10 13:30
11 14:00
12 14:30
13 15:00
14 15:30
15 16:00
16 16:30
17 17:00
18 17:30
19 18:00
I need to get the total sum hours, for example from 09:00 to 18:00 there is a total of :
9
hours, I need to get this sum of hours
Your table schema hour is varchar, you need to cast as time, then do the calculation
SELECT datediff(hour,min(cast(hour as time)),max(cast(hour as time)))
FROM Timetable
sqlfiddle
NOTE
I would suggest your hour column as datetime or time instead of varchar. because hour column intention is time.
EDIT
If your time is 9:00 to 17:30, you can try to use datediff minute to get the total diff minutes then divide 60 to get hours.
SELECT datediff(minute,min(cast(hour as time)),max(cast(hour as time))) / CAST(60 as float)
FROM Timetable
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=6e005cdfad4eca3ff7c4c92ef14cc9c7
use datediff function
select datediff(hour,min(h),max(h)) from
(
select CAST(hour AS TIME) as h from Timetable
) as t
strongly disagreed to put time value in varchar ,so it is better change your data type from varchar to time
declare #a time = '13:00',#b time = '17:30' --- Here you can give time, what you need.
select distinct convert(varchar(20)
, datediff(MINUTE,#a,#b) / 60)
+ ':' +
convert(varchar(20), datediff(MINUTE,#a,#b) % 60)
from #Timetable
where hour in (#a,#b)
For your SQL Fiddle Sample Data.
Obviously, you need to use datediff(). However, you should be doing the datediff() in minutes or seconds and then converting to hours:
SELECT datediff(minute, min(cast(hour as time)), max(cast(hour as time))) / 60.0
FROM Timetable;
This will handle the case where the number of hours is not an exact number of hours.