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;
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
将GROUP BY添加到要检查重复的字段 您的查询可能看起来像
SELECT field1, field2, field3, ...... FROM table GROUP BY field1
将检查Field1以排除重复记录
或者你可能会问
SELECT * FROM table GROUP BY field1
字段1的重复记录被排除在SELECT中
其他回答
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
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。
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。
这样只需要1个查询就可以得到2个唯一的列 select Distinct col1,col2 from '{path}' group by col1,col2 如果需要,可以增加列
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