How to compare two dates to find time difference in SQL Server 2005, date manipulation

Question

I have two columns:

job_start                         job_end
2011-11-02 12:20:37.247           2011-11-02 13:35:14.613

How would it be po

Answer 1

Cast the result as TIME and the result will be in time format for duration of the interval.

select CAST(job_end - job_start) AS TIME(0)) from tableA

Answer 2

If you trying to get worked hours with some accuracy, try this (tested in SQL Server 2016)

SELECT DATEDIFF(MINUTE,job_start, job_end)/60.00;

Various DATEDIFF functionalities are:

SELECT DATEDIFF(year,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(quarter,     '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(month,       '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(dayofyear,   '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(day,         '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(week,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(hour,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(minute,      '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(second,      '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(millisecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');

Ref: https://docs.microsoft.com/en-us/sql/t-sql/functions/datediff-transact-sql?view=sql-server-2017

Answer 3

Try this in Sql Server

SELECT 
      start_date as firstdate,end_date as seconddate
       ,cast(datediff(MI,start_date,end_date)as decimal(10,3)) as minutediff
      ,cast(cast(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) as int ) as varchar(10)) + ' ' + 'Days' + ' ' 
      + cast(cast((cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) - 
        floor(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60)) ) * 24 as int) as varchar(10)) + ':' 

     + cast( cast(((cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) 
      - floor(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60)))*24
        -
        cast(floor((cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60) 
      - floor(cast(datediff(MI,start_date,end_date)as decimal(10,3)) / (24*60)))*24) as decimal)) * 60 as int) as varchar(10))

    FROM [AdventureWorks2012].dbo.learndate

Answer 4

Below code gives in hh:mm format.

select RIGHT(LEFT(job_end- job_start,17),5)

Answer 5

I think you need the time gap between job_start & job_end.

Try this...

select SUBSTRING(CONVERT(VARCHAR(20),(job_end - job_start),120),12,8) from tableA

I ended up with this.

01:14:37

Answer 6

You can use the DATEDIFF function to get the difference in minutes, seconds, days etc.

SELECT DATEDIFF(MINUTE,job_start,job_end)

MINUTE obviously returns the difference in minutes, you can also use DAY, HOUR, SECOND, YEAR (see the books online link for the full list).

If you want to get fancy you can show this differently for example 75 minutes could be displayed like this: 01:15:00:0

Here is the code to do that for both SQL Server 2005 and 2008

-- SQL Server 2005
SELECT CONVERT(VARCHAR(10),DATEADD(MINUTE,DATEDIFF(MINUTE,job_start,job_end),'2011-01-01 00:00:00.000'),114)

-- SQL Server 2008
SELECT CAST(DATEADD(MINUTE,DATEDIFF(MINUTE,job_start,job_end),'2011-01-01 00:00:00.000') AS TIME)