Help with SQL. Pick off most recent records.

I have a table that contains, other other things, a job compete date/time column and a job ID column.  For for each job ID, I want to pull back the most recent record, using the job compete date/time column, so that I end up with one record per job ID.   Thanks
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HLRosenbergerAuthor Commented:
so I could so this for a specific job.  How do I make this work for all job IDs?

  select top 1 * from Audit where jobID =  'MT40'
  order by [job_datetime] desc
HLRosenbergerAuthor Commented:
And I also need all other columns in the table to be returned.
HLRosenbergerAuthor Commented:
Also, I could do this to get a list of all jobIDs.  

   select distinct(jobID) from [Audit]

now I just need to somehow combine with, for all jobIDs.

select top 1 * from Audit where jobID =  'MT40'
  order by [job_datetime] desc
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Scott PletcherSenior DBACommented:
SELECT col_name1, col_name2, ...
    FROM table_name
) AS derived
WHERE row_num = 1

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declare @au_id char( 11 )

set rowcount 0
select * into #mytemp from jobs

set rowcount 1

select @au_id = au_id from #mytemp

while @@rowcount <> 0
    set rowcount 0
    select top 1 * from Audit where jobID =  @au_id
  order by [job_datetime] desc
    delete #mytemp where au_id = @au_id

    set rowcount 1
    select @au_id = au_id from #mytemp
set rowcount 0
PortletPaulEE Topic AdvisorCommented:
Please use the ROW_NUMBER() approach as shown by Scott Pletcher.

The windowing functions are very efficient and you can avoid resorting to cursors.
HLRosenbergerAuthor Commented:
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Microsoft SQL Server

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