Hibernate通过@DynamicInsert注解实现动态插入不生效

Question

插入和更新，对于空字段，生成的sql不想包含改字段，查了资料说使用这个的，但是，我的却没效果，和没使用是一样的，部分代码：
import java.util.Date;
import java.util.Set;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.ManyToMany;
import javax.persistence.Table;

import org.hibernate.annotations.DynamicInsert;
import org.hibernate.annotations.DynamicUpdate;

/**

• 系统用户表 添加动态插入和动态更新
  */
  @Entity
  @Table(name = "user", catalog = "showlove")
  @DynamicInsert(true)
  @DynamicUpdate(true)
  public class User extends BaseModel {

  private static final long serialVersionUID = 1L;

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  @Column(name = "userId", nullable = false)
  private Integer userId;

  /**

  • 角色Id,级联更新、刷新，懒加载 */ @ManyToMany(fetch = FetchType.LAZY, cascade = {CascadeType.REFRESH,CascadeType.MERGE }) @JoinTable(name = "users_roles", joinColumns = @JoinColumn(name = "userId"), inverseJoinColumns = @JoinColumn(name = "roleId")) private Set roles;

  @Column(name = "username", length = 18, nullable = false)
  private String userName;

  @Column(name = "password", length = 18, nullable = false)
  private String password;

  @Column(name = "sex", length = 4, nullable = true)
  private String sex;

  @Column(name = "nickname", length = 16, nullable = true)
  ……
  测试代码：
  @Test
  public void testUpdate(){
  User user=new User();
  user.setUserId(24);
  // Role role=new Role();
  // role.setRoleId(1);
  user.setActivated(1);
  userService.update(user);
  }
  生成的sql：
  update
  showlove.user
  set
  activated=?,
  age=?,
  createDate=?,
  email=?,
  nickname=?,
  password=?,
  phone=?,
  sex=?,
  username=?
  where
  userId=?