I wish to convert the dates from each column to minutes, sum all the minutes and find the average time between each row. I have to fetch the data from a SQL database and display the final amount of minutes in a textbox.
Thanks for any help!
-Remi
Edit:
I solved this problem like this:
SELECT Datediff(minute, startdate, enddate) FROM «table-name»»)
Answer:
I solved this problem like this: SELECT Datediff(minute, startdate, enddate) FROM «table-name»»)
Related
I need to calculate the time differences on a given day. I tried something like that but not works.
CONVERT(TIME, DATEADD(MINUTE, DATEDIFF(MI, MIN(CreatedOn),
MAX(CreatedOn)), 0), 108) AS WorkingTime
Thanks guys
You can try this way, since you don't show the error you get I don't know what detail you can have, but to get the time difference you can get it like this as I leave you here below
DECLARE #MaxDate DATETIME=DATEADD(HOUR,20,GETDATE()) --here you could get the maximum of your table separately
DECLARE #MinDate DATETIME=GETDATE() --here you could get the minimum of your table separately
SELECT DATEDIFF(MINUTE,#MinDate,#MaxDate) AS WorkingTime
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=09128de0d996e85a21992197b524a208
Your question is not entirely clear, but I think this is what you're trying to do.
SELECT CAST(WorkingTime AS DATE),
MIN(WorkingTime) AS min,
MAX(WorkingTime) AS max,
DATEADD(HOUR, DATEDIFF(HOUR, 0, MIN(WorkingTime)), 0) AS truncatedmin,
DATEDIFF(MINUTE,DATEADD(HOUR, DATEDIFF(HOUR, 0, MIN(WorkingTime)), 0),MAX(WorkingTime)) AS [difference]
FROM YourTable
GROUP BY CAST(WorkingTime AS DATE)
The column called "difference" should show the number of minutes difference between the truncatedmin and max columns.
From this column I need to calculate the working time for each day, for this I need to select the first time on a given day and then round it to the full hour (06:08:20 to 06:00:00) and then calculate the differences between the last time in day, which will give me the working time in minutes.
My column
I need to group data in intervals of 15 (or X) minutes in sql.
For example, i have the next scenario.
The result that i expect to obtain is
I tried using Lag function but i dont get what i want, because it add interval to each row and continues grouping.
Thanks in advance and apologies for my bad english.
If you want the intervals to be calendar based -- i.e. four per hour starting at 0, 15, 30, and 45 minutes, then you can use:
select id, min(begin_date), max(begin_date)
from t
group by id, convert(date, begin_date),
datepart(hour, begin_date), datepart(minute, begin_date) / 15;
Note that begin date and end date have the same value, so I just used begin_date in this answer.
I am new to tableau and i need to find all row set filter on basis of given time frame and then grouping the result of same minute.I can crack this problem in Sql as following
select count(x)
from table t
where datediff(30, min, date)
group by datepart(min, date)
How to fix it in tableau ?
I'm not really sure of what you want but this will give you the number of record per minute, for the records whose date is in the last 30 minutes:
select datepart(MI, date) , COUNT(*) as NumberOfRecordsInMinute
from table t
where date >= dateADD(MI, -30, getdate())
group by datepart(MI, date)
order by datepart(MI, date)
It isn't obvious what your SQL is trying to do here as the standard SQL datediff function works like this datediff(datepart,startdate,enddate) and returns the difference between two dates in the units specified.
If that is what you want, then the Tableau function datediff does much the same job and datediff('minute',start date,enddate) will return the number of minutes between two datetimes and it will be grouped already because the result is discrete.
You can then count the number of records matching each minute difference.
i'm trying to count the total days between current date and a specific column called "DayConfirm" (datetime). I want to show the total days in a new column beside the rows with "DayChanged" So far i got this:
SELECT DATEDIFF(CURDATE(),(DayConfirm, '%m/%d/%Y') AS DAYS
FROM Administr
can anyone help me? Thank you in advance.
You could try to make use of DATEDIFF:
SELECT DATEDIFF(DAY, GETDATE() , DayConfirm)
Your need to pass Date Part ie day here attribute:
SELECT DATEDIFF(day,'2008-06-05','2008-08-05') AS DiffDate
also you missing the closing ); try this:
SELECT DATEDIFF(day, CURDATE(),(DayConfirm, '%m/%d/%Y')) AS DAYS
FROM Administr
please help me with my problem. So, I have a table named 'RATES' which contains these columns:
id (int)
rate (money)
start_time (datetime)
end_time(datetime)
example data:
1 150 8:00am 6:00pm
2 200 6:00pm 4:00am
3 250 8:00am 4:00am (the next day)
What I have to do is to select all the id(s) to where a given time would fall.
e.g given time: 9:00 pm, the output should be 2,3
The problem is I got this time range between 8am to 4am the next day and I don't know what to do. Help, please! thanks in advance :D
Assuming that #Andriy M is correct:
Data never spans more than 24 hours
if end_time<=start_time then end_time belongs to the next day
then what you're looking for is this:
Declare #GivenTime DateTime
Set #GivenTime = '9:00 PM'
Select ID
From Rates
Where (Start_Time<End_Time And Start_Time<=#GivenTime And End_Time>=#GivenTime)
Or (Start_Time=End_Time And Start_Time=#GivenTime)
Or (Start_Time>End_Time And (Start_Time>=#GivenTime Or End_Time<=#GivenTime))
I don't really ever use MS SQL, but maybe this will help.
I was going to suggest something like this, but by the way you have your data set up, this would fail.
SELECT id FROM RATES
WHERE datepart(hh, start_time) <= 9 AND datepart(hh, end_time) >= 9;
You'll have you search using the actual date if you expect to get the correct data back.
SELECT id FROM RATES
WHERE start_time <= '2011-1-1 9:00' AND end_time >= '2011-1-1 9:00';
This may not be exactly correct, but it may help you look in the right direction.
I guess #gbn is not going to help you. I will try and fill in.
Given -- a table called timedata that has ranges only going over at most one day
WITH normalized AS
(
SELECT *
FROM timedata
WHERE datepart(day,start_time) = datepart(day,endtime)
UNION ALL
SELECT id, rate, start_time, dateadd(second,dateadd(day,datediff(day,0,end_time),0),-1) as end_time
FROM timedata
WHERE not (datepart(day,start_time) = datepart(day,endtime))
UNION ALL
SELECT id, rate,dateadd(day,datediff(day,0,end_time),0) as start_time, end_time
FROM timedata
WHERE not (datepart(day,start_time) = datepart(day,endtime))
)
SELECT *
FROM normalized
WHERE datepart(hour,start_time) < #inhour
AND datepart(hour,end_time) > #inhour
This makes use of a CTE and a trick to truncate datetime values. To understand this trick read this question and answer: Floor a date in SQL server
Here is an outline of what this query does:
Create a normalized table with each time span only going over one day by
Selecting all rows that occur on the same day.
Then for each entry that spans two days joining in
Selecting the starttime and one second before the next day as the end time for all that span.
and
Selecting 12am of the end_time date as the starttime and the end_time.
Finally you perform the select using the hour indicator on this normalized table.
If your ranges go over more than one day you would need to use a recursive CTE to get the same normalized table.