这样
UPDATE HT_USER_LANGUAGE SET HT_USER_LANGUAGE.`LEARNLANG1` = 5,HT_USER_LANGUAGE.`SKILLLEVEL1`= 4 WHERE USERID = 105745提示
查询:UPDATE HT_USER_LANGUAGE SET HT_USER_LANGUAGE.`LEARNLANG1` = 5,HT_USER_LANGUAGE.`SKILLLEVEL1`= 4 WHERE USERID = 105745 错误代码: 1054
Unknown column 'SKILLLEVEL1' in 'field list'
但是
SELECT SKILLLEVEL1 FROM HT_USER_LANGUAGE WHERE USERID = 105745;
结果出来了!求解mysql
-> SET HT_USER_LANGUAGE.`LEARNLANG1` = 5,HT_USER_LANGUAGE.`SKILLLEVEL1`= 4
-> WHERE USERID = 105745;
Query OK, 0 rows affected (0.00 sec)
Rows matched: 0 Changed: 0 Warnings: 0
列一定是存在的,不然
SELECT SKILLLEVEL1 FROM HT_USER_LANGUAGE WHERE USERID = 105745;
也应该报错啊,后面创建了一个一样的表2insert into select 导入原来表1的数据
在删除老表1,修改表2名为表1。又可以了怀疑是不是表结构破坏了,或者名字有冲突还是怎么样
贴出以供分析。贴出来看一下不就知道了?然后把 show create table HT_USER_LANGUAGE; 也贴出来。
*************************** 1. row ***************************
Table: HT_USER_LANGUAGE
Create Table: CREATE TABLE `HT_USER_LANGUAGE` (
`USERID` int(11) unsigned NOT NULL COMMENT '鐢ㄦ埛ID',
`NATIVELANG` tinyint(4) unsigned DEFAULT '0' COMMENT '姣嶈 璇█璇浣跨
敤鏁板瓧浠h〃',
`LEARNLANG1` tinyint(4) unsigned DEFAULT '0' COMMENT '瑕佸涔犵殑璇█1',
`SKILLLEVEL1` tinyint(4) unsigned DEFAULT '0' COMMENT '璇█1鐨勭啛缁冪瓑绾?, `LEARNLANG2` tinyint(4) unsigned DEFAULT '0',
`SKILLLEVEL2` tinyint(4) unsigned DEFAULT '0',
`LEARNLANG3` tinyint(4) unsigned DEFAULT '0',
`SKILLLEVEL3` tinyint(4) unsigned DEFAULT '0',
`LEARNLANG4` tinyint(4) unsigned DEFAULT '0',
`SKILLLEVEL4` tinyint(4) unsigned DEFAULT '0',
PRIMARY KEY (`USERID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
1 row in set (0.00 sec)ERROR:
No query specified