小样样520 2015-08-17 12:04 采纳率: 55.6%
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hql 级联更新问题 报不知道什么错

dao层代码:
Query query = createQuery("update SfdfUserPublisher set sfdfUser.pwd='123' where id='402881f04f2bfdc9014f2c0029f50000'");
return query.executeUpdate()>0
报错信息:
Hibernate: update sfdf_user_publisher cross join set PWD='123' where ID='402881f04f2bfdc9014f2c0029f50000'
19:46:07,615 WARN JDBCExceptionReporter:233 - SQL Error: 1064, SQLState: 42000
19:46:07,615 ERROR JDBCExceptionReporter:234 - 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 'set PWD='123' where ID='402881f04f2bfdc9014f2c0029f50000'' at line 1
方法名:com.sfdf.controller.SfdfUserAction.addOrEditPublisher(SfdfUserAction.java:222) &操作:新增发布者信息 &结果:失败
19:46:07,616 ERROR SfdfUserAction:47 - 方法名:com.sfdf.controller.SfdfUserAction.addOrEditPublisher(SfdfUserAction.java:222) &操作:新增发布者信息 &结果:失败
org.hibernate.exception.SQLGrammarException: could not execute update query
model层代码:
@ManyToOne(fetch = FetchType.LAZY,cascade=CascadeType.ALL)
@JoinColumn(name = "USER_ID", nullable = false)
public SfdfUser getSfdfUser() {
return this.sfdfUser;
}

public void setSfdfUser(SfdfUser sfdfUser) {
    this.sfdfUser = sfdfUser;
}

不知道是什么错,求解救
  • 写回答

3条回答 默认 最新

  • devmiao 2015-08-17 22:46
    关注

    MySQL server version for the right syntax to use near 'set PWD='123' where ID='402881f04f2bfdc9014f2c0029f50000'' at line 1

    sql语法错,你的类型对不对

    评论

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