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Change Value based on Dynamical Object RRS feed

  • Question

  • Goal:
    The goal is to make the variable with value '2014-06-14 09:00:00.000'

    Problem:
    The syntax code is created as a dynamical object how do you make it from '2014-06-14 16:20:10.000' into value '2014-06-14 09:00:00.000'?

    DECLARE @a datetime = '2014-06-14 16:20:10.000'

    Monday, June 16, 2014 9:19 PM

Answers

  • Are you looking for this?

    DECLARE @a datetime = '2014-06-14 16:20:10.000'
    
    select dateadd(hour,9,convert(datetime,convert(date,@a)))
    
    set @a = '2014-06-14 08:20:10.000'
    select dateadd(hour,9,convert(datetime,convert(date,@a)))


    Satheesh
    My Blog | How to ask questions in technical forum


    • Proposed as answer by Elvis Long Friday, June 20, 2014 5:42 AM
    • Marked as answer by Elvis Long Thursday, June 26, 2014 1:26 AM
    Tuesday, June 17, 2014 7:02 AM
  • DECLARE @a datetime = '2014-06-14 16:20:10.000' -- pass any date value here
    
    SELECT DATEADD(dd,DATEDIFF(dd,0,@a),'09:00')


    Please Mark This As Answer if it helps to solve the issue Visakh ---------------------------- http://visakhm.blogspot.com/ https://www.facebook.com/VmBlogs

    • Proposed as answer by Elvis Long Friday, June 20, 2014 5:42 AM
    • Marked as answer by Elvis Long Thursday, June 26, 2014 1:26 AM
    Tuesday, June 17, 2014 7:08 AM

All replies

  • The syntax code is created as a dynamical object how do you make it from '2014-06-14 16:20:10.000' into value '2014-06-14 09:00:00.000'?

    Hello,

    Can you explain the logic for this, please? 


    Olaf Helper

    [ Blog] [ Xing] [ MVP]

    Tuesday, June 17, 2014 6:35 AM
  • Are you looking for this?

    DECLARE @a datetime = '2014-06-14 16:20:10.000'
    
    select dateadd(hour,9,convert(datetime,convert(date,@a)))
    
    set @a = '2014-06-14 08:20:10.000'
    select dateadd(hour,9,convert(datetime,convert(date,@a)))


    Satheesh
    My Blog | How to ask questions in technical forum


    • Proposed as answer by Elvis Long Friday, June 20, 2014 5:42 AM
    • Marked as answer by Elvis Long Thursday, June 26, 2014 1:26 AM
    Tuesday, June 17, 2014 7:02 AM
  • DECLARE @a datetime = '2014-06-14 16:20:10.000' -- pass any date value here
    
    SELECT DATEADD(dd,DATEDIFF(dd,0,@a),'09:00')


    Please Mark This As Answer if it helps to solve the issue Visakh ---------------------------- http://visakhm.blogspot.com/ https://www.facebook.com/VmBlogs

    • Proposed as answer by Elvis Long Friday, June 20, 2014 5:42 AM
    • Marked as answer by Elvis Long Thursday, June 26, 2014 1:26 AM
    Tuesday, June 17, 2014 7:08 AM