I have a SQL query returning a value for x, which is a timestamp, mapped to a C# object of type long:
SELECT DATEDIFF(second, { d '1970-01-01'}, dateCompleted) AS x
The above statement works. However, I need to get the timestamp to return the value in milliseconds rather than seconds. In SQL Server 2016 I can do this:
SELECT DATEDIFF_BIG(millisecond, { d '1970-01-01'}, dateCompleted) AS x
...and that works great. However, I'm stuck on SQL Server 2008 R2.
I could return the values and do some post-processing in C# to multiply x by 1000 but I wondered if there's a way to handle this in the query itself. I've tried a simple multiplication but that yields an Arithmetic overflow error:
SELECT DATEDIFF(second, { d '1970-01-01'}, dateCompleted) * 1000 AS x
Could anyone suggest how to accomplish this?
Thanks.
DATEDIFF returns an INT so it cannot be used to return difference in millisecond if the two dates are far (approx. 25 days) apart. However you could calculate the difference in seconds, BIGINT multiply by 1000, and add the milliseconds:
SELECT DATEDIFF(SECOND, '1970-01-01', dateCompleted)
* CAST(1000 AS BIGINT)
+ DATEPART(MILLISECOND, dateCompleted)
Assuming you want UNIX timestamp you also need to add the timezone offset to the result (I hope you stored it along with date completed).
How about using cast() or convert()?
SELECT DATEDIFF(second,{ d '1970-01-01'},dateCompleted) * convert(bigint, 1000) AS x
Perhaps a variable?
DECLARE #milli BIGINT;
SET #milli = DATEDIFF(second,{ d '1970-01-01'},dateCompleted) * 1000.0;
SELECT #milli;
Datediff return int value so second will have issue with int data type.
you can get the minutes or days and multiple with 60 for getting seconds
SELECT DATEDIFF(m,{ d '1970-01-01'},getdate()) * 1000 * 60
Related
I have a Table in SQL server with a column "Time" having data type as time(7). Need to call the sum of this column, and when I use the following statement, it returns result as integer only.
Eg. If total time is 1:30:00,I expect result as 1.5. But the code I use doesn't get me this, it get me result as 1. Please check if you have a solution.
The code I used is
SELECT SUM(DATEPART(ss,Time) + DATEPART(mi,Time)*60 + DATEPART(hh,Time)*3600)/3600 AS TotalTime FROM dbo.Table
SELECT (
DATEPART(hh,Time) +
DATEPART(mi,Time) / 60.0 +
DATEPART(ss,Time) / 3600.0
) AS TotalTime
FROM dbo.Table
Try below - you don't need sum() function here and in your case, it is showing 1 because your result is 5400/3600 which is 1 but you need to add a float value as you are expecting float result
SELECT (DATEPART(ss,'1:30:00') + DATEPART(mi,'1:30:00')*60 +
DATEPART(hh,'1:30:00')*3600)/3600.00
AS TotalTime FROM dbo.Table
Try this, you can change the datepart argument based on your needs here is the full list
SELECT SUM(CAST(DATEDIFF(MINUTE, '00:00:00', [Time]) as float)/60) AS TotalHours FROM [dbo].[Table]
When you divide some value by int type, the result will be also int (the fraction is just dropped). Therefore, you need to convert a divider of 3600 from int to decimal:
SELECT SUM(DATEPART(ss,Time) + DATEPART(mi,Time)*60 + DATEPART(hh,Time)*3600)/CONVERT(DECIMAL(16,4), 3600) AS TotalTime FROM dbo.Table
If you want the difference in decimal hours, then do the following:
Convert the time values to seconds.
Sum the seconds.
Divide by 60 * 60
So:
select sum(datediff(second, 0, v.t)) / (60.0 * 60)
from (values (convert(time, '00:10:01')),
(convert(time, '01:00:03'))
) v(t)
There is no reason to break the value in to component parts. That just seems unnecessarily complicated.
I am being supplied a single integer that is supposed to represent an hour. So If it returns 1 it is 1:00 am and so forth on a 24 hour clock,13 for example is 1:00 pm. I need to convert this into time in SQL.
I know in MYSQL they have a function which does this:
SEC_TO_TIME(TheHour*60*60)
Is there an equivalent I can use in SQL? How do I do this?
You could do something like this.
select cast(DATEADD(hour, 13, 0) as time)
The upside is that it will still work even with negative numbers or values over 24.
There are two T-SQL function:
DATEFROMPARTS ( year, month, day )
and
TIMEFROMPARTS ( hour, minute, seconds, fractions, precision )
then you can use CONVERT if you need to format it.
-- test data
declare #hour_table table(hour_number int)
while (select count(*) from #hour_table) < 24
begin
insert into #hour_table(hour_number)
select count(*) from #hour_table
end
-- return results with your conversion to time string
select
hour_number,
convert(varchar(8),timefromparts( hour_number, 0, 0, 0, 0 ),0) as time_string
from #hour_table
I m using this query to get a result of the difference between the start time and end time of an activity. Where the end time is null i wanted to put the minimum value as 500. Please advice and HELP!!
select * from table
where (end_time - start_time) * 24 * 60 > 1,
IF end_time IS NULL THEN '500';
So this is your query:
select * from table where (end_time - start_time) * 24 * 60 > 1;
But you want to treat a null end_time as 500. So use NVL or COALESCE to replace the null with 500:
select * from table where (nvl(end_time,500) - start_time) * 24 * 60 > 1;
IF end_time IS NULL THEN '500';
Just to make it more clear, '500' is not a number rather a string since it is enclosed within single quotation marks.
Now, end_time is. DATE data type or a timestamp, ideally. So, 500 makes no sense. You must convert it to appropriate type, whether 500 is days, hours, minutes, seconds, fraction of a second.
As in other answer it is suggested to use NVL(end_time, 500), it makes no sense. What does 500 - a date mean? Applying NVL is the need, however, you must convert it to the required value, else those are two different data types and Oracle won't allow it.
UPDATE
In my opinion,
Difference between two dates gives the number of days to the precision of seconds converted back to days. But, difference between an arbitrary number and a date makes no sense.
I assumed that start_time and end_time columns have number as datatype, for this calculation you need to select these specific columns and not all (*). Comparison is in where clause, this works in oracle11.
select ((NVL(END_TIME, 500)-START_TIME) * 24 * 60) from TABLE_NAME where ((NVL(END_TIME, 500)-START_TIME) * 24 * 60) > 1;
I am trying to do write in a function to subtract a few dates, then divide them together, add 1 then * 100 to get an overall % complete.
((SYSDATETIME() - proj.proj_scheduledDate)/(proj.proj_dueDate - proj.proj_scheduledDate + 1)) * 100 as 'percent done
returns
Operand data type datetime is invalid for divide operator.
What do I need to do to get this query to work? THANKS!
Might I suggest using DATEDIFF() with dd for a day datepart -- you can easily get the percentage of days completed - or go to smaller datepart values if you need to.
Returns the count (signed integer) of the specified datepart
boundaries crossed between the specified startdate and enddate.
SELECT
FLOOR((DATEDIFF(NOW(), proj.proj_scheduledDate) / DATEDIFF(proj.proj_dueDate, proj.proj_scheduledDate)) * 100)
FROM
foo;
I got a column called DateOfBirth in my csv file with Excel Date Serial Number Date
Example:
36464
37104
35412
When i formatted cells in excel these are converted as
36464 => 1/11/1999
37104 => 1/08/2001
35412 => 13/12/1996
I need to do this transformation in SSIS or in SQL. How can this be achieved?
In SQL:
select dateadd(d,36464,'1899-12-30')
-- or thanks to rcdmk
select CAST(36464 - 2 as SmallDateTime)
In SSIS, see here
http://msdn.microsoft.com/en-us/library/ms141719.aspx
The marked answer is not working fine, please change the date to "1899-12-30" instead of "1899-12-31".
select dateadd(d,36464,'1899-12-30')
You can cast it to a SQL SMALLDATETIME:
CAST(36464 - 2 as SMALLDATETIME)
MS SQL Server counts its dates from 01/01/1900 and Excel from 12/30/1899 = 2 days less.
tldr:
select cast(#Input - 2e as datetime)
Explanation:
Excel stores datetimes as a floating point number that represents elapsed time since the beginning of the 20th century, and SQL Server can readily cast between floats and datetimes in the same manner. The difference between Excel and SQL server's conversion of this number to datetimes is 2 days (as of 1900-03-01, that is). Using a literal of 2e for this difference informs SQL Server to implicitly convert other datatypes to floats for very input-friendly and simple queries:
select
cast('43861.875433912' - 2e as datetime) as ExcelToSql, -- even varchar works!
cast(cast('2020-01-31 21:00:37.490' as datetime) + 2e as float) as SqlToExcel
-- Results:
-- ExcelToSql SqlToExcel
-- 2020-01-31 21:00:37.490 43861.875433912
this actually worked for me
dateadd(mi,CONVERT(numeric(17,5),41869.166666666664)*1440,'1899-12-30')
(minus 1 more day in the date)
referring to the negative commented post
SSIS Solution
The DT_DATE data type is implemented using an 8-byte floating-point number. Days are represented by whole number increments, starting with 30 December 1899, and midnight as time zero. Hour values are expressed as the absolute value of the fractional part of the number. However, a floating point value cannot represent all real values; therefore, there are limits on the range of dates that can be presented in DT_DATE. Read more
From the description above you can see that you can convert these values implicitly when mapping them to a DT_DATE Column after converting it to a 8-byte floating-point number DT_R8.
Use a derived column transformation to convert this column to 8-byte floating-point number:
(DT_R8)[dateColumn]
Then map it to a DT_DATE column
Or cast it twice:
(DT_DATE)(DT_R8)[dateColumn]
You can check my full answer here:
Is there a better way to parse [Integer].[Integer] style dates in SSIS?
Found this topic helpful so much so created a quick SQL UDF for it.
CREATE FUNCTION dbo.ConvertExcelSerialDateToSQL
(
#serial INT
)
RETURNS DATETIME
AS
BEGIN
DECLARE #dt AS DATETIME
SELECT #dt =
CASE
WHEN #serial is not null THEN CAST(#serial - 2 AS DATETIME)
ELSE NULL
END
RETURN #dt
END
GO
I had to take this to the next level because my Excel dates also had times, so I had values like this:
42039.46406 --> 02/04/2015 11:08 AM
42002.37709 --> 12/29/2014 09:03 AM
42032.61869 --> 01/28/2015 02:50 PM
(also, to complicate it a little more, my numeric value with decimal was saved as an NVARCHAR)
The SQL I used to make this conversion is:
SELECT DATEADD(SECOND, (
CONVERT(FLOAT, t.ColumnName) -
FLOOR(CONVERT(FLOAT, t.ColumnName))
) * 86400,
DATEADD(DAY, CONVERT(FLOAT, t.ColumnName), '1899-12-30')
)
In postgresql, you can use the following syntax:
SELECT ((DATE('1899-12-30') + INTERVAL '1 day' * FLOOR(38242.7711805556)) + (INTERVAL '1 sec' * (38242.7711805556 - FLOOR(38242.7711805556)) * 3600 * 24)) as date
In this case, 38242.7711805556 represents 2004-09-12 18:30:30 in excel format
In addition of #Nick.McDermaid answer I would like to post this solution, which convert not only the day but also the hours, minutes and seconds:
SELECT DATEADD(s, (42948.123 - FLOOR(42948.123))*3600*24, dateadd(d, FLOOR(42948.123),'1899-12-30'))
For example
42948.123 to 2017-08-01 02:57:07.000
42818.7166666667 to 2017-03-24 17:12:00.000
You can do this if you just need to display the date in a view:
CAST will be faster than CONVERT if you have a large amount of data, also remember to subtract (2) from the excel date:
CAST(CAST(CAST([Column_With_Date]-2 AS INT)AS smalldatetime) AS DATE)
If you need to update the column to show a date you can either update through a join (self join if necessary) or simply try the following:
You may not need to cast the excel date as INT but since the table I was working with was a varchar I had to do that manipulation first. I also did not want the "time" element so I needed to remove that element with the final cast as "date."
UPDATE [Table_with_Date]
SET [Column_With_Excel_Date] = CAST(CAST(CAST([Column_With_Excel_Date]-2 AS INT)AS smalldatetime) AS DATE)
If you are unsure of what you would like to do with this test and re-test! Make a copy of your table if you need. You can always create a view!
Google BigQuery solution
Standard SQL
Select Date, DATETIME_ADD(DATETIME(xy, xm, xd, 0, 0, 0), INTERVAL xonlyseconds SECOND) xaxsa
from (
Select Date, EXTRACT(YEAR FROM xonlydate) xy, EXTRACT(MONTH FROM xonlydate) xm, EXTRACT(DAY FROM xonlydate) xd, xonlyseconds
From (
Select Date
, DATE_ADD(DATE '1899-12-30', INTERVAL cast(FLOOR(cast(Date as FLOAT64)) as INT64) DAY ) xonlydate
, cast(FLOOR( ( cast(Date as FLOAT64) - cast(FLOOR( cast(Date as FLOAT64)) as INT64) ) * 86400 ) as INT64) xonlyseconds
FROM (Select '43168.682974537034' Date) -- 09.03.2018 16:23:28
) xx1
)
For those looking how to do this in excel (outside of formatting to a date field) you can do this by using the Text function https://exceljet.net/excel-functions/excel-text-function
i.e.
A1 = 132134
=Text(A1,"MM-DD-YYYY") will result in a date
This worked for me because sometimes the field was a numeric to get the time portion.
Command:
dateadd(mi,CONVERT(numeric(17,5),41869.166666666664)*1440,'1899-12-31')