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
sql;">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。
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