Question

我正在尝试计算列中值的出现次数，并返回出现次数大于1的行。

例如：

SomeId   AnotherId
1        BC4
1        CDE
1        SDF
5        ER
5        WER
4        DFGR4

我希望能够获得列SomeId中出现次数大于1的SomeId。

我看过使用SUM和GROUP，但没有运气:(

Answer 1

这将返回多次出现的ID。

SELECT SomeId
FROM yourTable
GROUP BY SomeId
HAVING COUNT(1) > 1

Answer 2

SELECT SomeId
    FROM YourTable
    GROUP BY SomeId
    HAVING COUNT(SomeId) > 1;

Answer 3

SELECT SomeId, COUNT(SomeId) AS someidcount
FROM tablename
GROUP BY SomeId
HAVING someidcount > 1

Answer 4

试试这个：

select count(*) as frequency, SomeId group by SomeId having frequency > 1

Answer 5

您使用GROUP BY走在正确的轨道上，但您还需要HAVING。

试试这个：

SELECT SomeId, COUNT(SomeId)
FROM table
GROUP BY SomeId
HAVING COUNT(SomeId) > 1

以下是初学者a Wikipedia article on the HAVING clause的链接。

Answer 6

试试这个：

select SomeId, Count(distinct AnotherId)
  from YourTable
 group by SomeId
having Count(distinct AnotherId) > 1

Answer 7

如果您熟悉PHP，可以这样做：

$sql = "SELECT * FROM table";
$res = mysql_query($sql);
while($row = mysql_fetch_assoc):
if($row['SomeId'] > 1):

{do whatever you like here...}

endif;
endwhile;

返回mysql中具有重复列值的记录

7 个答案: