SQL: How To Select Earliest Row

Question

I have a report that looks something like this:

CompanyA      Workflow27     June5
CompanyA      Workflow27     June8
CompanyA      Workflow27     June12
CompanyB      Workflow13     Apr4
CompanyB      Workflow13     Apr9
CompanyB      Workflow20     Dec11
CompanyB      Wofkflow20     Dec17

This is done with SQL (specifically, T-SQL version Server 2005):

SELECT company
   , workflow
   , date
FROM workflowTable

I would like the report to show just the earliest dates for each workflow:

CompanyA      Workflow27     June5
CompanyB      Workflow13     Apr4
CompanyB      Workflow20     Dec11

Any ideas? I can't figure this out. I've tried using a nested select that returns the earliest tray date, and then setting that in the WHERE clause. This works great if there were only one company:

SELECT company
   , workflow
   , date
FROM workflowTable
WHERE date = (SELECT TOP 1 date
              FROM workflowTable
              ORDER BY date)

but this obviously won't work if there is more than one company in that table. Any help is appreciated!

Answer 1

Simply use min()

SELECT company, workflow, MIN(date) 
FROM workflowTable 
GROUP BY company, workflow

Answer 2

In this case a relatively simple GROUP BY can work, but in general, when there are additional columns where you can't order by but you want them from the particular row which they are associated with, you can either join back to the detail using all the parts of the key or use OVER():

Runnable example (Wofkflow20 error in original data corrected)

;WITH partitioned AS (
    SELECT company
        ,workflow
        ,date
        ,other_columns
        ,ROW_NUMBER() OVER(PARTITION BY company, workflow
                            ORDER BY date) AS seq
    FROM workflowTable
)
SELECT *
FROM partitioned WHERE seq = 1

Answer 3

SELECT company
   , workflow
   , MIN(date)
FROM workflowTable
GROUP BY company
       , workflow