I am having a couple of issues converting time to decimal...
Essentially I have used a DATEDIFF to identify the hours and minutes between two dates and times. This produces a result of 5.45, however I would like to get the result as 5.75.
select RTRIM(DATEDIFF(second, startdate, enddate)/3600)+'.'
+ RIGHT('0'+RTRIM((DATEDIFF(second, startdate, enddate) % 3600)/60),2)
from time
I have tried a few things but I believe the issue is that this is not an integer and that's why i cant convert the minutes.
Get the time difference in minutes, and divide by 60, this will place hours before the decimal separator and minutes after:
datediff(minute, startdate, enddate) / 60.0
Select DATEDIFF(second, startdate, enddate)/3600.0
As mentioned in this answer, "Take the DateDiff in seconds instead, and then divide by 86400.0. The decimal point is required."
Related
I have two variables 'triage_date_time' and 'checkin_date_time'. Both are formatted as, for example, 2018-12-31 14:13:00. Showing the year-month-day and hour-minute-second both within one cell.
I wanted to create a variable that calculates the time it takes from check-in to triage.
I attempted to use the following code:
SELECT DISTINCT datediff(minute, 'triage_date_time', 'checkin_date_time') as
checkin_to_triage
However, when running this code I get the following error... "Conversion failed when converting date and/or time from character string".
Any suggestions of how I can write a code that would calculate the minute difference of these two variables.
Thanks!
One problem is obviously the single quotes. Assuming that you are using SQL Server, variables start with #. So:
select datediff(minute, #triage_date_time, #checkin_date_time) as checkin_to_triage
If you are confused and really mean columns in a table, then:
select datediff(minute, triage_date_time, checkin_date_time) as checkin_to_triage
from t;
could it be that your field is a CHARACTER data type ?
cast your char to datetime
SELECT DISTINCT datediff(minute, CAST(triage_date_time AS datetime), CAST(checkin_date_time AS datetime)) as checkin_to_triage
Try with this query
DECLARE #triage_date_time DATETIME = '20181231 14:13:00'
DECLARE #checkin_date_time DATETIME = '20181231 16:13:00'
SELECT DATEDIFF (MINUTE, #triage_date_time, #checkin_date_time) AS 'checkin_to_triage'
Output :
checkin_to_triage
120
I have a function which is working fine in MySQL
round((now()-ts/60) as tdiff
(round the result of subtracting the current datetime from ts (also a datetime) divided by 60)
Attempting (and failing) to convert this for SQL Server.
Tried -
round((GETDATE()-ts/60) as tdiff
but that results in round function requires 2 or 3 parameters (which to me it does), so modified to -
round((GETDATE()-ts/60,0) as tdiff
but that results in the datatypes (GETDATE and ts) are incompatible in the subtract operator.
So then I attempted to cast both GETDATE and ts as date and that made no difference.
ts is a conventional datetime i.e.
2918-04-20 11:05:09 and I assumed GETDATE returned the same format.
As an example if GETDATE is today and ts is 2018-04-20 11:05:09 then tdiff is
6850891 (round effectively removes the dashes and colons and concatenates the datetime producing 20180420110509 for 2018-04-20 11:05:09 and 20180831164000 for 2018-08-31 16:40:00 and then divides by 60 to get 6850891.
Is there a remedy for this?
Regards, Ralph
GETDATE(), as per the documentation, returns a datetime. A datetime is accurate to 1/300 of a second, and it's accuracy cannot be changed.
If you want the time accurate to a second, you need to convert to a datetime2(0):
SELECT CONVERT(datetime2(0),GETDATE());
Also, however, don't use syntax like GETDATE()-ts. use the functions DATEADD and DATEDIFF for date maths.
I've no idea what GETDATE()-ts/60 is trying to acheive. Perhaps the number of minutes between the 2? DATEDIFF counts the "ticks" between 2 dates/times, thus DATEDIFF(MINUTE,'00:00:59','00:01:00') would return 1, despite there only being 1 second between the 2 times. This is because the minute value has "ticked" once (from 0 to 1). Therefore you might want to use DATEDIFF(SECOND,'00:00:59','00:01:00') / 60. This would return 0, as 1 / 60 in integer math is 0 (as is 59 / 60).
I think you want to use the DATEDIFF function:
DATEDIFF ( datepart , startdate , enddate )
DATEDIFF ( second, ts, GETDATE())
DATEDIFF ( second, ts, GETDATE())
DATEDIFF ( minute, ts, GETDATE())
DATEDIFF ( hour, ts, GETDATE())
The first argument tells it which increment of time to return.
If you are trying to find the difference between two values, then use datediff(). For instance:
select datediff(day, getdate(), ts)
gets the difference in days.
date_diff() or a related function would also be the right approach in MySQL.
sorry, I don't know if I have understand the question, you need to do date-date/60 and round the result?
In this case you have to change the minus ("-") with the DATEDIFF("Type return example DAYS", GETDATE(), ts).
So you will have ROUND((DATEDIFF(DAY,GETDATE(),ts)/60,0)
Please try and let me know if it will works for you
Bye
I've been stumbling with this issue for a couple days now, and cannot seem to figure out why, when my getdate() insert into the columns are providing a millisecond decimal to the military time format, I still cannot seem to be able to pull a decimal format datediff() result. Does it have to do with the engine not recognizing the decimal due to the surrounding '' characters?
When I use:
select datediff(s,'2013-06-01 21:59:59.141','2013-06-01 23:59:59.997')
It returns:
7200
And when I use:
select cast(datediff(s,'2013-06-01 21:59:59.141','2013-06-01 23:59:59.997') as float);
It returns:
7200
I am at a loss as to what I am missing in order to result in a decimal value.
Thanks
If you are trying to get the milliseconds of the difference, and you want to convert the units to seconds, you can try using something like the following:
SELECT DATEDIFF(MS,'2013-06-01 21:59:59.141','2013-06-01 23:59:59.997') / 1000.0
That'll produce: 7200.856000.
Please note that DATEDIFF(MS, ...) requires guarding for long time spans or it will give an overflow:
SELECT datediff(MS, '2013-06-30 23:59:59.997', '2013-06-01 21:59:59.141')
-- FAILURE: The datediff function resulted in an overflow. The number of dateparts separating two date/time instances is too large
You can use following method which is overflow-safe and gives you a float result:
SELECT cast(cast('2013-06-01 23:59:59.997' as datetime)-cast('2013-06-01 21:59:59.141' as datetime) as float) * 24.0
-- Returned 2.00023796296296
SELECT cast(cast('2013-06-30 23:59:59.997' as datetime)-cast('2013-06-01 21:59:59.141' as datetime) as float) * 24.0
-- Returned 698.000237962963
I'm not sure how this method works when time zone changed during the measured date period.
I am trying to add some hours in SQL SERVER using DATEADD function. But when I try this,
SELECT DATEADD(Hour, 0.5, GETDATE())
It is not adding 0.5 hour. How to solve this?
You can't. It's well describer on documentation: DATEADD (Transact-SQL)
number
Is an expression that can be resolved to an int that is added
to a datepart of date. User-defined variables are valid.
If you specify a value with a decimal fraction, the fraction is truncated and
not rounded.
UPDATE
You could try that:
SELECT DATEADD(Second, 0.5 * 60 * 60, GETDATE())
Of course - you can change DATEPART and multiplier to get desired precision.
You can't add parts of hours, just full hours. Use minute for half an hour
SELECT DATEADD(minute, 30, GETDATE())
I have found another approach to this where adding a number to the date instead of using the function works for fractions. For example:
GETDATE() + n
where n is the number of days. For 1 half hour, you can use:
GETDATE() + 0.5/24
I know that when i do the following, it converts getdate to int
select cast (getdate() as int)
Getdate output on my server is "2010-06-11 14:42:20.100" and the int to which the above command is converting to is 40339. What is this integer? Did this int consider minutes and
seconds? i am confused. Please help.
Regards
Manjot
This number is the number of days since the reference date, which is 01/01/1900. You should get the same result from:
SELECT DATEDIFF(day, 0, GETDATE())
Internally, SQL stores datetimes as two 4-byte integers. The first integer (which you are getting in your statement) represents the number of days since the reference date. The second integer represents the number of 1/300 second intervals since midnight. You can find the second integer by converting first to binary, then to int:
SELECT CONVERT(int, CONVERT(binary(4), GETDATE()))