Question

Name      Age      Cut-off
--------------------------
Ram       22        89.50
Ganesh    22        66.00
Sam       22        92.00
Albert    22        89.50
Kannan    22        65.45   
John      22        66.00

在上表中，Ram和Albert以及Ganesh和John具有相同的截止值。怎么能我得到所有这些行？

Answer 1

SELECT  a.*
FROM    tableName a
        INNER JOIN
        (
            SELECT `Cut-off`, COUNT(*) totalCOunt
            FROM    tableName
            GROUP   BY `Cut-off`
            HAVING  COUNT(*) > 1
        ) b ON a.`Cut-off` = b.`Cut-off`

要获得更快的效果，请在列INDEX

上添加Cut-off

SQLFiddle Demo

Answer 2

这是一种方法：

select *
from t
where exists (select 1 from t t2 where t2.cutoff = t.cutoff and t2.name <> t.name)

如何获得具有相同截止标记的行集？

2 个答案: