关于避免MySQL替换逻辑SQL的坑爹操作详解
replace into和insert into on duplicate key 区别
replace的用法
当不冲突时相当于insert,其余列默认值
当key冲突时,自增列更新,replace冲突列,其余列默认值
Com_replace会加1
Innodb_rows_updated会加1
Insert into …on duplicate key的用法
不冲突时相当于insert,其余列默认值
当与key冲突时,只update相应字段值。
Com_insert会加1
Innodb_rows_inserted会增加1
实验展示
表结构
create table helei1( id int(10) unsigned NOT NULL AUTO_INCREMENT, name varchar(20) NOT NULL DEFAULT '', age tinyint(3) unsigned NOT NULL default 0, PRIMARY KEY(id), UNIQUE KEY uk_name (name) ) ENGINE=innodb AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;
表数据
root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 1 | 贺磊 | 26 | | 2 | 小明 | 28 | | 3 | 小红 | 26 | +----+-----------+-----+ 3 rows in set (0.00 sec)
replace into用法
root@127.0.0.1 (helei)> replace into helei1 (name) values('贺磊'); Query OK, 2 rows affected (0.00 sec) root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 0 | +----+-----------+-----+ 3 rows in set (0.00 sec) root@127.0.0.1 (helei)> replace into helei1 (name) values('爱璇'); Query OK, 1 row affected (0.00 sec) root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 0 | | 5 | 爱璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec)
replace的用法
当没有key冲突时,replace into 相当于insert,其余列默认值
当key冲突时,自增列更新,replace冲突列,其余列默认值
Insert into …on duplicate key:
root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 0 | | 5 | 爱璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) root@127.0.0.1 (helei)> insert into helei1 (name,age) values('贺磊',0) on duplicate key update age=100; Query OK, 2 rows affected (0.00 sec) root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 100 | | 5 | 爱璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 100 | | 5 | 爱璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) root@127.0.0.1 (helei)> insert into helei1 (name) values('爱璇') on duplicate key update age=120; Query OK, 2 rows affected (0.01 sec) root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 100 | | 5 | 爱璇 | 120 | +----+-----------+-----+ 4 rows in set (0.00 sec) root@127.0.0.1 (helei)> insert into helei1 (name) values('不存在') on duplicate key update age=80; Query OK, 1 row affected (0.00 sec) root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小红 | 26 | | 4 | 贺磊 | 100 | | 5 | 爱璇 | 120 | | 8 | 不存在 | 0 | +----+-----------+-----+ 5 rows in set (0.00 sec)
总结
replace into这种用法,相当于如果发现冲突键,先做一个delete操作,再做一个insert 操作,未指定的列使用默认值,这种情况会导致自增主键产生变化,如果表中存在外键或者业务逻辑上依赖主键,那么会出现异常。因此建议使用Insert into …on duplicate key。由于编写时间也很仓促,文中难免会出现一些错误或者不准确的地方,不妥之处恳请读者批评指正。
好了,以上就是这篇文章的全部内容了,希望本文的内容对大家的学习或者工作具有一定的参考学习价值,如果有疑问大家可以留言交流,谢谢大家对我们的支持。
赞 (0)