mysql重复记录查询
1.一个字段重复，如果记录多于2天，最多只显示2条

2.sql语句

SQL code

CREATE TABLE `teacher` (  `ID` bigint(20) NOT NULL COMMENT 'ID',  `REGNAME` char(200) DEFAULT NULL COMMENT '师教编号',  `NAME` char(20) NOT NULL COMMENT '教师名称',  `TYPE` int(1) DEFAULT '0' COMMENT '师教类型（0：内聘，1：外聘）',  PRIMARY KEY (`ID`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;INSERT INTO `teacher` VALUES ('10001', 'allen', 'allen', '0');INSERT INTO `teacher` VALUES ('10002', 'ruby', 'ruby', '0');INSERT INTO `teacher` VALUES ('10003', 'sharon', 'sharon', '1');INSERT INTO `teacher` VALUES ('10004', 'alpha', 'alpha', '0');INSERT INTO `teacher` VALUES ('10005', 'alpha', 'alpha', '0');INSERT INTO `teacher` VALUES ('10006', 'alpha', 'alpha', '0');

1.数据库记录

2.需要的结果


[解决办法]
SELECT * FROM `teacher`;
SELECT * FROM `teacher` A WHERE 2>=(SELECT COUNT(*) FROM `teacher`
WHERE A.`NAME`=`NAME` AND A.`ID`>=`ID`

)

[解决办法]
在Oracle中可以这样
查询所有有重复字段的所有记录

select * from teacher A where (select count(*) from teacher where A.NAME=NAME )>=2;

而后查询前两条记录
SELECT * FROM (select * from teacher A where (select count(*) from teacher where A.NAME=NAME )>=2) E1 WHERE
(SELECT COUNT(*) FROM (select * from teacher A where (select count(*) from teacher where A.NAME=NAME )>=2) E2 WHERE E1.NAME=E2.NAME AND E2.ID<E1.ID )<2 order by NAME,ID;
[解决办法]
这样也可以的：

SQL code

SELECT a.id,a.regname,a.name,a.type FROM teacher a LEFT JOIN teacher b ON b.name=a.name AND b.id>a.id GROUP BY a.id,a.regname,a.name,a.type HAVING COUNT(b.id) < 2;
[解决办法]
SQL code
mysql> select * from teacher;+-------+---------+--------+------+| ID    | REGNAME | NAME   | TYPE |+-------+---------+--------+------+| 10001 | allen   | allen  |    0 || 10002 | ruby    | ruby   |    0 || 10003 | sharon  | sharon |    1 || 10004 | alpha   | alpha  |    0 || 10005 | alpha   | alpha  |    0 || 10006 | alpha   | alpha  |    0 |+-------+---------+--------+------+6 rows in set (0.00 sec)mysql> select * from teacher t    -> where 2>(select count(*) from teacher where REGNAME=t.REGNAME and ID<t.ID);+-------+---------+--------+------+| ID    | REGNAME | NAME   | TYPE |+-------+---------+--------+------+| 10001 | allen   | allen  |    0 || 10002 | ruby    | ruby   |    0 || 10003 | sharon  | sharon |    1 || 10004 | alpha   | alpha  |    0 || 10005 | alpha   | alpha  |    0 |+-------+---------+--------+------+5 rows in set (0.02 sec)mysql>
[解决办法]
SELECT A.NAME,COUNT(*) FROM `teacher` A  
LEFT JOIN `teacher` B
ON A.`NAME`=B.`NAME` AND A.`ID`>=B.`ID`
看看结果