这是一个简单的解决方法:

 WITH cte AS /* Declaring a new table named 'cte' to be a clone of your table */
 (SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY val1 DESC) AS rn
 FROM MyTable /* Selecting only unique values based on the "id" field */
 )
 SELECT * /* Here you can specify several columns to retrieve */
 FROM cte
 WHERE rn = 1

SELECT  c2.field1 ,
        field2
FROM    (SELECT DISTINCT
                field1
         FROM   dbo.TABLE AS C
        ) AS c1
        JOIN dbo.TABLE AS c2 ON c1.field1 = c2.field1

SELECT * from table where field in (SELECT distinct field from table)

它可以通过内部查询来完成

$query = "SELECT * 
            FROM (SELECT field
                FROM table
                ORDER BY id DESC) as rows               
            GROUP BY field";

对于SQL Server，您可以使用dense_rank和其他窗口函数来获取指定列上具有重复值的所有行和列。这里有一个例子……

with t as (
    select col1 = 'a', col2 = 'b', col3 = 'c', other = 'r1' union all
    select col1 = 'c', col2 = 'b', col3 = 'a', other = 'r2' union all
    select col1 = 'a', col2 = 'b', col3 = 'c', other = 'r3' union all
    select col1 = 'a', col2 = 'b', col3 = 'c', other = 'r4' union all
    select col1 = 'c', col2 = 'b', col3 = 'a', other = 'r5' union all
    select col1 = 'a', col2 = 'a', col3 = 'a', other = 'r6'
), tdr as (
    select 
        *, 
        total_dr_rows = count(*) over(partition by dr)
    from (
        select 
            *, 
            dr = dense_rank() over(order by col1, col2, col3),
            dr_rn = row_number() over(partition by col1, col2, col3 order by other)
        from 
            t
    ) x
)

select * from tdr where total_dr_rows > 1

这是对col1、col2和col3的每个不同组合进行行计数。

好问题@aryaxt——你可以看出这是一个好问题，因为你5年前问过这个问题，而我今天在试图找到答案时偶然发现了它!

我只是试图编辑接受的答案，以包括这一点，但如果我的编辑没有使它:

如果你的表不是那么大，并且假设你的主键是一个自动递增的整数，你可以这样做:

SELECT 
  table.*
FROM table
--be able to take out dupes later
LEFT JOIN (
  SELECT field, MAX(id) as id
  FROM table
  GROUP BY field
) as noDupes on noDupes.id = table.id
WHERE
  //this will result in only the last instance being seen
  noDupes.id is not NULL