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

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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

[email protected] (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

[email protected] (helei)> replace into helei1 (name) values('贺磊');
Query OK, 2 rows affected (0.00 sec)

[email protected] (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)
[email protected] (helei)> replace into helei1 (name) values('爱璇');
Query OK, 1 row affected (0.00 sec)

[email protected] (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:

[email protected] (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)

[email protected] (helei)> insert into helei1 (name,age) values('贺磊',0) on duplicate key update age=100;
Query OK, 2 rows affected (0.00 sec)

[email protected] (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)

[email protected] (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)

[email protected] (helei)> insert into helei1 (name) values('爱璇') on duplicate key update age=120;
Query OK, 2 rows affected (0.01 sec)

[email protected] (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)

[email protected] (helei)> insert into helei1 (name) values('不存在') on duplicate key update age=80;
Query OK, 1 row affected (0.00 sec)

[email protected] (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

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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。