You have an error in your SQL syntax; check the manual that corresponds to your
hibernate进行查询的时候报错：queryString=from SysUser as model where model.usrName=?adminHibernate: select sysuser0_.usr_id as usr1_7_, sysuser0_.usr_role_id as usr2_7_, sysuser0_.usr_name as usr3_7_, sysuser0_.usr_password as usr4_7_, sysuser0_.usr_flag as usr5_7_ from db_crm.dbo.sys_user sysuser0_ where sysuser0_.usr_name=?org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute queryCaused by: org.hibernate.exception.SQLGrammarException: could not execute query at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:90)Caused by: com.mysql.jdbc.exceptions.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 '.sys_user sysuser0_ where sysuser0_.usr_name='admin'' at line 1
查询代码如下：
public List findByProperty(String propertyName, Object value) {
log.debug("finding SysUser instance with property: " + propertyName
+ ", value: " + value);
try {
String queryString = "from SysUser as model where model."
+ propertyName + "=?";
System.out.println("queryString="+queryString+value);
return getHibernateTemplate().find(queryString, value);
} catch (RuntimeException re) {
log.error("find by property name failed", re);
throw re;
}
} MySQL --------------------编程问答-------------------- select sysuser0_.usr_id as usr1_7_, sysuser0_.usr_role_id as usr2_7_, sysuser0_.usr_name as usr3_7_, sysuser0_.usr_password as usr4_7_, sysuser0_.usr_flag as usr5_7_ from db_crm.dbo.sys_user sysuser0_ where sysuser0_.usr_name=?
这个语句在数据库里能查到东西？ --------------------编程问答-------------------- SQL语句有问题。。你把SQL语句打印出来 然后放数据库里执行调试吧，成功了再移植到程序里来 --------------------编程问答-------------------- 谢谢。。。。。 --------------------编程问答-------------------- 你写的hql有问题吧。。。。
补充：Java , Java EE
