Question

所以我有下表：

mysql> show create table user_api_skills \G
*************************** 1. row ***************************
       Table: user_api_skills
Create Table: CREATE TABLE `user_api_skills` (
  `characterID` int(11) NOT NULL,
  `typeID` int(11) NOT NULL,
  `level` enum('0','1','2','3','4','5') NOT NULL DEFAULT '0',
  `skillpoints` int(11) NOT NULL,
  `currentTime` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`characterID`,`typeID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
1 row in set (0.00 sec)

mysql>

在该表中，我试图插入/更新一行：

mysql> SELECT * FROM `user_api_skills` WHERE `characterID` =93192782 AND `typeID` =3359;
+-------------+--------+-------+-------------+---------------------+
| characterID | typeID | level | skillpoints | currentTime         |
+-------------+--------+-------+-------------+---------------------+
|    93192782 |   3359 | 3     |      135765 | 2013-09-30 16:58:35 |
+-------------+--------+-------+-------------+---------------------+
1 row in set (0.00 sec)

我相信我的查询是正确形成的，并且在执行时它不会抛出任何错误或警告：

mysql> INSERT INTO user_api_skills (characterID,typeID,level,skillpoints)
VALUES (93192782,3359,4,135765) ON DUPLICATE KEY UPDATE level=4,             
skillpoints=135765,currentTime=NOW();
Query OK, 2 rows affected (0.22 sec)

我更新了2行（正如我期望的那样，在dup update上插入）

mysql> SELECT * FROM `user_api_skills` WHERE `characterID` =93192782 AND `typeID` =3359;
+-------------+--------+-------+-------------+---------------------+
| characterID | typeID | level | skillpoints | currentTime         |
+-------------+--------+-------+-------------+---------------------+
|    93192782 |   3359 | 3     |      135765 | 2013-09-30 16:59:13 |
+-------------+--------+-------+-------------+---------------------+
1 row in set (0.00 sec)

mysql>

但行本身只更改单个值（currentTime）。任何人都可以解释为什么其他两个字段没有更新？

Answer 1

对不起，我自己解决了这个问题。级别字段是ENUM，查询将新值指定为数字。将查询更新为以下内容会产生预期结果。

mysql> INSERT INTO user_api_skills (characterID,typeID,level,skillpoints) VALUES 
(93192782,3359,4,135765) ON DUPLICATE KEY UPDATE level='4', skillpoints=135765, 
currentTime=NOW();

通过为更新提供一个int，它更新为enum的一个基数选择，所以在这个例子中，第四个选择是'3'。

MySQL INSERT ... ON DUPLICATE KEY没有更新表而没有错误或警告

1 个答案: