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

你可以用with子句来实现。

例如:

WITH c AS (SELECT DISTINCT a, b, c FROM tableName)
SELECT * FROM tableName r, c WHERE c.rowid=r.rowid AND c.a=r.a AND c.b=r.b AND c.c=r.c

这还允许您只选择WITH子句查询中选择的行。

其他回答

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

Try

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

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。

你可以用with子句来实现。

例如:

WITH c AS (SELECT DISTINCT a, b, c FROM tableName)
SELECT * FROM tableName r, c WHERE c.rowid=r.rowid AND c.a=r.a AND c.b=r.b AND c.c=r.c

这还允许您只选择WITH子句查询中选择的行。

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