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這篇文章給大家分享的是有關(guān)如何避免MySQL替換邏輯SQL的坑的內(nèi)容。小編覺(jué)得挺實(shí)用的,因此分享給大家做個(gè)參考,一起跟隨小編過(guò)來(lái)看看吧。
replace into和insert into on duplicate key 區(qū)別
replace的用法
當(dāng)不沖突時(shí)相當(dāng)于insert,其余列默認(rèn)值
當(dāng)key沖突時(shí),自增列更新,replace沖突列,其余列默認(rèn)值
Com_replace會(huì)加1
Innodb_rows_updated會(huì)加1
Insert into …on duplicate key的用法
不沖突時(shí)相當(dāng)于insert,其余列默認(rèn)值
當(dāng)與key沖突時(shí),只update相應(yīng)字段值。
Com_insert會(huì)加1
Innodb_rows_inserted會(huì)增加1
實(shí)驗(yàn)展示
表結(jié)構(gòu)
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;
表數(shù)據(jù)
root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 1 | 賀磊 | 26 | | 2 | 小明 | 28 | | 3 | 小紅 | 26 | +----+-----------+-----+ 3 rows in set (0.00 sec)
replace into用法
root@127.0.0.1 (helei)> replace into helei1 (name) values('賀磊'); 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 | 小紅 | 26 | | 4 | 賀磊 | 0 | +----+-----------+-----+ 3 rows in set (0.00 sec) root@127.0.0.1 (helei)> replace into helei1 (name) values('愛(ài)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 | 小紅 | 26 | | 4 | 賀磊 | 0 | | 5 | 愛(ài)璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec)
replace的用法
當(dāng)沒(méi)有key沖突時(shí),replace into 相當(dāng)于insert,其余列默認(rèn)值
當(dāng)key沖突時(shí),自增列更新,replace沖突列,其余列默認(rèn)值
Insert into …on duplicate key:
root@127.0.0.1 (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小紅 | 26 | | 4 | 賀磊 | 0 | | 5 | 愛(ài)璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) root@127.0.0.1 (helei)> insert into helei1 (name,age) values('賀磊',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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛(ài)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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛(ài)璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) root@127.0.0.1 (helei)> insert into helei1 (name) values('愛(ài)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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛(ài)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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛(ài)璇 | 120 | | 8 | 不存在 | 0 | +----+-----------+-----+ 5 rows in set (0.00 sec)
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