關於避免MySQL替換邏輯SQL的坑爹操作詳解
阿新 • • 發佈:2020-01-09
replace into和insert into on duplicate key 區別
replace的用法
當不衝突時相當於insert,其餘列預設值
當key衝突時,自增列更新,replace衝突列,其餘列預設值
Com_replace會加1
Innodb_rows_updated會加1
Insert into …on duplicate key的用法
不衝突時相當於insert,其餘列預設值
當與key衝突時,只update相應欄位值。
Com_insert會加1
Innodb_rows_inserted會增加1
實驗展示
表結構
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;
表資料
[email protected] (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
[email protected] (helei)> replace into helei1 (name) values('賀磊'); Query OK,2 rows affected (0.00 sec) [email protected] (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) [email protected] (helei)> replace into helei1 (name) values('愛璇'); Query OK,1 row affected (0.00 sec) [email protected] (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小紅 | 26 | | 4 | 賀磊 | 0 | | 5 | 愛璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec)
replace的用法
當沒有key衝突時,replace into 相當於insert,其餘列預設值
當key衝突時,自增列更新,replace衝突列,其餘列預設值
Insert into …on duplicate key:
[email protected] (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小紅 | 26 | | 4 | 賀磊 | 0 | | 5 | 愛璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) [email protected] (helei)> insert into helei1 (name,age) values('賀磊',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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) [email protected] (helei)> select * from helei1; +----+-----------+-----+ | id | name | age | +----+-----------+-----+ | 2 | 小明 | 28 | | 3 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛璇 | 0 | +----+-----------+-----+ 4 rows in set (0.00 sec) [email protected] (helei)> insert into helei1 (name) values('愛璇') 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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛璇 | 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 | 小紅 | 26 | | 4 | 賀磊 | 100 | | 5 | 愛璇 | 120 | | 8 | 不存在 | 0 | +----+-----------+-----+ 5 rows in set (0.00 sec)
總結
replace into這種用法,相當於如果發現衝突鍵,先做一個delete操作,再做一個insert 操作,未指定的列使用預設值,這種情況會導致自增主鍵產生變化,如果表中存在外來鍵或者業務邏輯上依賴主鍵,那麼會出現異常。因此建議使用Insert into …on duplicate key。由於編寫時間也很倉促,文中難免會出現一些錯誤或者不準確的地方,不妥之處懇請讀者批評指正。
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