SQL236. 删除emp_no重复的记录,只保留最小的id对应的记录。
描述
删除emp_no重复的记录,只保留最小的id对应的记录。 id | emp_no | title | from_date | to_date |
1 | 10001 | Senior Engineer | 1986-06-26 | 9999-01-01 |
2 | 10002 | Staff | 1996-08-03 | 9999-01-01 |
3 | 10003 | Senior Engineer | 1995-12-03 | 9999-01-01 |
4 | 10004 | Senior Engineer | 1995-12-03 | 9999-01-01 |
示例1
输入:
drop table if exists titles_test; CREATE TABLE titles_test ( id int(11) not null primary key, emp_no int(11) NOT NULL, title varchar(50) NOT NULL, from_date date NOT NULL, to_date date DEFAULT NULL); insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'), ('2', '10002', 'Staff', '1996-08-03', '9999-01-01'), ('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'), ('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'), ('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'), ('6', '10002', 'Staff', '1996-08-03', '9999-01-01'), ('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');
输出:
1|10001|Senior Engineer|1986-06-26|9999-01-01 2|10002|Staff|1996-08-03|9999-01-01 3|10003|Senior Engineer|1995-12-03|9999-01-01 4|10004|Senior Engineer|1995-12-03|9999-01-01
Sqlite 解法, 执行用时: 9ms, 内存消耗: 3304KB, 提交时间: 2021-06-08
delete from titles_test where id not in(select * from (select min(id)from titles_test group by emp_no)as a);
Sqlite 解法, 执行用时: 9ms, 内存消耗: 3304KB, 提交时间: 2021-06-04
delete from titles_test where id not in (select min (id)from titles_test group by emp_no)
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3200KB, 提交时间: 2020-07-07
DELETE FROM titles_test WHERE id NOT IN (SELECT MIN(id) FROM titles_test GROUP BY emp_no)
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3276KB, 提交时间: 2020-11-10
/*方法一*/ delete from titles_test where id not in (select min(id) from titles_test group by emp_no)
Sqlite 解法, 执行用时: 10ms, 内存消耗: 3296KB, 提交时间: 2021-07-16
delete from titles_test where id not in ( select min(id) from titles_test group by emp_no )