SELECT DISTINCT field1, field2, field3, ......
FROM table;

我试图完成以下SQL语句,但我希望它返回所有列。 这可能吗?

就像这样:

SELECT DISTINCT field1, * 
FROM table;

当前回答

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

其他回答

我建议使用

SELECT  * from table where field1 in 
(
  select distinct field1 from table
)

这样,如果field1在多个行中有相同的值,将返回所有记录。

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";
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

Try

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