SQL Server: How do I return all columns where one is not repeated?

Question

I've got a table with repeating IDs which I will fix later. Basically I want to return all the rows where the ID is distinct, but I want the ENTIRE row. Something like:

select * from table group by ID

select * from table where (ID is not repeated)

In this case, they are identical rows, so I don't care if it's First or Last, Min or Max.

Note that I DON"T want to do this:

select MIN(col1), MIN(col2), ... from table group by ID

I want a way to get this result without enumerating every column.

Edit: I'm using SQL Server 2008 R2.

Answer 1

If you are using MySql, do this:

select 
    *
from tbl
group by ID

MySQL live test: http://www.sqlfiddle.com/#!2/8c7fd/2

If you are using Postgresql, do this:

select distinct on(id)
    *
from tbl
order by id

If you want Postgresql DISTINCT ON be at least as predictable as CTE windowing function. Sort another column:

select distinct on(id)
    *
from tbl
order by id
   , someColumnHere -- Choose ASC for first row, DESC for last row

Postgresql live test: http://www.sqlfiddle.com/#!1/8c7fd/1

If you are using CTE windowing-capable database (e.g. Postgres, Oracle, Sql Server), use this:

with ranked as
(
  select 
      rank() over(partition by id order by column) rn,
      *
  from tbl
)
select * from ranked where rn = 1

CTE windowing-capable databases:

Posgtresql: http://www.sqlfiddle.com/#!1/8c7fd/2

Oracle: http://www.sqlfiddle.com/#!4/b5cf9/1

Sql Server: http://www.sqlfiddle.com/#!3/8c7fd/3