How to add back a time component to a date

Hello,

I'm working on creating a view.  IThe underlying table has many fields with four fields of note
StartDate - Datetime
StartHour - Decimal
DueDate - Datetime
DueHour - Decimal

When the database was designed, it appears that instead of a datetime field containing both the time and date components, only the date data was entered, so dates would appear and be stored as 2011-08-26 00:00:00:0000.  The time was entered as a decimal number so for example a time of 23:30 would be entered as 23.5
I want to add the corresponding time back to the date field so that I can calculate the corresponding time difference in minutes between the (StartDate + Startime) and (DueDate + DueTime) via the datediff function.  I can already convert the decimal time component to hh:mm:ss via CONVERT(VARCHAR(8), Dateadd(minute, [duehour]*60,0),108) AS HMS.  I'm just not sure how to add it to the date time component as a calculated field so that it would appear as a datetime 2011-08-26 23:30:00:000.  This database is running on SQL Server 2005
Juan VelasquezAsked:
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Juan VelasquezAuthor Commented:
I think I may have figured it out
[duedate] + CONVERT(DateTime, Dateadd(minute, [duehour]*60,0),108)  AS NewDueDate
Leo TorresSQL DeveloperCommented:
Not sure how you mean add

but it should be similar to this
Select dateadd(hour,3,getdate())

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Juan VelasquezAuthor Commented:
Here is the sql string that I came up with
SELECT 
      [jobnum]   
      ,[opcode]              
      ,[partnum]
      ,[description]
      ,[vendornum]
	  ,[startdate]
	  ,[starthour]     
      ,[duedate]
      ,[duehour]	
	  ,DATEDIFF(minute,[startdate] + CONVERT(DateTime, Dateadd(minute, [starthour]*60,0),108),[duedate] + CONVERT(DateTime, Dateadd(minute, [duehour]*60,0),108)) AS 'Capacity'
      ,[opdesc]
      ,[PROGRESS_RECID]
      ,[PROGRESS_RECID_IDENT_]
  FROM [mfgtest803].[dbo].[joboper] 

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Leo TorresSQL DeveloperCommented:
I guess my question is how much time are you adding and what are you adding to? That why I can do the sql math.
Juan VelasquezAuthor Commented:
I appreciate the help.  Since the start time is a datetime datatype and the converted start hour is now also a date time datatype, I was able to add them.  In this case I thing I can use either the sql date add or the + operator
PortletPaulEE Topic AdvisorCommented:
Instead of the zero as third parameter  in this:

      Dateadd(minute, [duehour]*60,0)

use the relevant date column

      Dateadd(minute, [StartHour]*60,[StartDate])

      Dateadd(minute, [duehour]*60,[DueDate])

{+edit}
try to avoid using varchar conversions for date calculations, using varchars is slower than date functions

+edit2 and the datediff would be like this.  

DATEDEFF(minute,  Dateadd(minute, [StartHour]*60,[StartDate]) ,  Dateadd(minute, [duehour]*60,[DueDate]) )

but you may need to reverse the order of those 2 dates

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Juan VelasquezAuthor Commented:
Thanks for the help.  I will implement your suggestion
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Microsoft SQL Server 2005

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