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MS SQL SERVER  get countdown

Posted on 2011-02-25
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Last Modified: 2012-05-11
So whenever the user presses a button, it runs a sproc ... i need to return a value of current_date- top of the hour...so for example...if the user presses the submit at 1:31

Then it would return "29"..because its 29 minutes till 2:00
Next_delevery_date_go_hour = (top of the hour) - (currdate) .....

 DECLARE @CURR_DATE smalldatetime
 SET  @CURR_DATE = ISNULL(@CURR_DATE, getDATE())
-- i may not have this right...

--round it up...
 DECLARE @round_up_DATE smalldatetime
 SET  @round_up_DATE = ISNULL@round_up_DATE, getDATE())

declare @countdown nvarchar(10)
set  @countdown  = @round_up_DATE  - @CURR_DATE

--- i know this is off ....
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Question by:GlobaLevel
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4 Comments
 
LVL 4

Expert Comment

by:rocky_lotus_newbie
ID: 34982169
select (60-RIGHT(convert(varchar(5),getdate(),114),2))

This gives the number of minutes left till the next hour.

For Example, If you run this at 1.28 PM, it returns 32.

Thanks,
0
 
LVL 23

Expert Comment

by:wdosanjos
ID: 34982440
Alternatively:

SELECT 60 - DATEPART(MINUTE,GETDATE())


0
 
LVL 23

Expert Comment

by:wdosanjos
ID: 34982539
Here is the complete code.  It also correctly returns '0' at the hour (1:00 PM, returns 0).

DECLARE @COUNTDOWN NVARCHAR(10)

SET @COUNTDOWN = 60 - DATEPART(MINUTE,GETDATE())

IF @COUNTDOWN = '60' SET @COUNTDOWN = 0

Open in new window

0
 
LVL 17

Accepted Solution

by:
dbaSQL earned 2000 total points
ID: 34982553
Here you go:

DECLARE @pressbutton DATETIME = '2011-02-25 13:31:00.000'
DECLARE @topofhour DATETIME = DATEADD(mi, ROUND(DATEDIFF(mi, 0, @pressbutton) / 60.0, 0) * 60,0)

SELECT @pressbutton
SELECT @topofhour
SELECT DATEDIFF(mi,@pressbutton,@topofhour)
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