Question

我用hibernate的save方法插入数据报错，求大神指点

User [id=null, type=1, name=aa, code=123456789, company=aaa, tel=13811111, ip=127.0.0.1, insert_time=1415424295191, check_time=null, password=123, check=0]
Hibernate:
insert
into
tb_user
(type, name, code, company, tel, ip, insert_time, check_time, password, check)
values
(?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
org.hibernate.exception.SQLGrammarException: could not insert: [com.web.model.User]
at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:92)
at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:66)

Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'check) values ('1', 'aa', '123456789', 'aaa', '13811111', '127.0.0.1', 141542429' at line 1

Answer 1

你用的是注解还是配置，很明显你的id这一列的值是null,如果是配置把id的那一列改成identity

Answer 2

很明显其中一个字段有问题 需要我远程给你解决下吗

Answer 3

应该是类型转换的问题吧？或者check字段有问题？

追问

我把字段名全改了，结果报
Data truncation: Out of range value for column 'user_insert_time' at row 1
这个错误 mysql数据库中的时间我用int存的
java类中用long
这样好像不行，怎么改啊

追答

inserttime不是日期型的吗，你用日期啊