显示具有多行的键的所有行

Question

Table kal
id integer primary key
init char 4 indexed
job char4

id  init job
--+----+------
1 | aa | job1
2 | aa | job2
3 | bb | job1
4 | cc | job3   
5 | cc | job5

我想显示init有多行的所有行：

id  init job
--+----+------
1 | aa | job1
2 | aa | job2
4 | cc | job3   
5 | cc | job5

我试过了：

select * from kal where init in (select init from kal group by init having 
count(init)>2);

实际上，该表有60000行，查询是

count(init)<40, 

但是需要花费大量的时间，phpmyadmin和我的耐心耗尽。

两个

select init from kal group by init having count(init)>2)

和

select * from kal where init in ('aa','bb','cc')

在“没有时间”的情况下运行，不到0.02秒。

我尝试过不同的子查询，但都需要“无限”时间，超过几分钟;我实际上从来没有让他们完成。

Answer 1

你试过EXISTS吗？

select *
from yourtable t1
where exists (select count(*)
              from yourtable t2
              where t1.init = t2.init
              group by t2.init
              having count(*) >= 2);

请参阅SQL Fiddle with Demo

甚至这个：

select *
from yourtable t1
left join
(
  select count(*) cnt, init
  from yourtable
  group by init
) t2
  on t1.init = t2.init
where t2.cnt > 1

请参阅SQL Fiddle with Demo

Answer 2

以下是一个示例，您可以see it采取行动：

<强>查询

SELECT a.id, a.init, a.job
FROM kal a
INNER JOIN
  (SELECT init, COUNT(init) 
  FROM kal
  GROUP BY init
  HAVING COUNT(init) > 1) b ON b.init = a.init

<强>结果

| ID | INIT |  JOB |
--------------------
|  1 |   aa | job1 |
|  2 |   aa | job2 |
|  4 |   cc | job3 |
|  5 |   cc | job5 |