T-SQL: Deleting all duplicate rows but keeping one [duplicate]

Question

You didn't say what version you were using, but in SQL 2005 and above, you can use a common table expression with the OVER Clause. It goes a little something like this:

WITH cte AS (
  SELECT[foo], [bar], 
     row_number() OVER(PARTITION BY foo, bar ORDER BY baz) AS [rn]
  FROM TABLE
)
DELETE cte WHERE [rn] > 1

Play around with it and see what you get.

(Edit: In an attempt to be helpful, someone edited the ORDER BY clause within the CTE. To be clear, you can order by anything you want here, it needn't be one of the columns returned by the cte. In fact, a common use-case here is that "foo, bar" are the group identifier and "baz" is some sort of time stamp. In order to keep the latest, you'd do ORDER BY baz desc)

Example query:

DELETE FROM Table
WHERE ID NOT IN
(
SELECT MIN(ID)
FROM Table
GROUP BY Field1, Field2, Field3, ...
)

Here fields are column on which you want to group the duplicate rows.

Here's my twist on it, with a runnable example. Note this will only work in the situation where Id is unique, and you have duplicate values in other columns.

DECLARE @SampleData AS TABLE (Id int, Duplicate varchar(20))

INSERT INTO @SampleData
SELECT 1, 'ABC' UNION ALL
SELECT 2, 'ABC' UNION ALL
SELECT 3, 'LMN' UNION ALL
SELECT 4, 'XYZ' UNION ALL
SELECT 5, 'XYZ'

DELETE FROM @SampleData WHERE Id IN (
    SELECT Id FROM (
        SELECT 
            Id
            ,ROW_NUMBER() OVER (PARTITION BY [Duplicate] ORDER BY Id) AS [ItemNumber]
            -- Change the partition columns to include the ones that make the row distinct
        FROM 
            @SampleData
    ) a WHERE ItemNumber > 1 -- Keep only the first unique item
)

SELECT * FROM @SampleData

And the results:

Id          Duplicate
----------- ---------
1           ABC
3           LMN
4           XYZ

Not sure why that's what I thought of first... definitely not the simplest way to go but it works.