select min(table.id), table.column1
from table 
group by table.column1

Try

SELECT table.* FROM table 
WHERE otherField = 'otherValue'
GROUP BY table.fieldWantedToBeDistinct
limit x

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

SELECT *
FROM tblname
GROUP BY duplicate_values
ORDER BY ex.VISITED_ON DESC
LIMIT 0 , 30

在ORDER BY我刚刚把例子放在这里，你也可以在这里添加ID字段

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

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";