Spring + Hibernate应用程序

Question

我还有一个应用程序(Spring + Hibernate)和AppFuse框架。

我有两个实体：User和Robot，有一对一的关系.

我需要将下拉列表添加到Robot表单(robotForm.jsp)的所有者(robotForm.jsp)中。

Robot实体有一个User。我读到我必须为User创建一个自定义的User。(UserCustomEditor extends PropertyEditorSupport)和覆盖RobotFormController中的referenceData --在initBinder中添加。

RobotFormController

protected void initBinder(HttpServletRequest request, ServletRequestDataBinder binder) {
  SimpleDateFormat dateFormat = new SimpleDateFormat(getText("date.format"));
  dateFormat.setLenient(false);
  binder.registerCustomEditor(Date.class, null, 
  new CustomDateEditor(dateFormat, true));
  binder.registerCustomEditor(Long.class, null,
        new CustomNumberEditor(Long.class, null, true));
  binder.registerCustomEditor(User.class, new UserEditor(userManager));
}

// ...
protected Map referenceData(HtppServletRequest request) throws Exception {
  Map ownersMap = new HashMap();
  ownersMap.put("owners", userManager.getUsers();
  return ownersMap;
}

userManager.getUsers();返回用户列表。

UserEditor (也许这是我的错误)。

public class UserEditor extends PropertyEditorSupport {

  private final UserManager userManager;
  protected final transient Log log = LogFactory.getLog(getClass());

  public UserEditor(UserManager userManager) throws IllegalArgumentException {
    this.userManager = userManager;
  }

  @Override
  public void setAsText(String text) throws IllegalArgumentException {
   if (text != null && text.length() > 0) {
     try {
          User user = userManager.getUser(new String (text));
                    super.setValue(user);
         } catch (NumberFormatException ex) {
                    throw new IllegalArgumentException();
         }
      } else {
            super.setValue(null);
            }
     }

  @Override
  public String getAsText() {
    User user = (User) super.getValue();
    return  (user != null ? (user.getId()+"").toString(): "");
  }

} 

robotForm.jsp

<form:select path="owner" itemValue="id" itemLabel="name" items="${owners}"
</form:select>  

我在NullPointerException方法的ownersMap.put("owners", userManager.getUsers();行中得到了一个referenceData。

编辑：

UserManagerImpl

@Service(value = "userManager")
public class UserManagerImpl implements UserManager {
@Autowired
UserDao dao;

public void setUserDao(UserDao dao) {
    this.dao = dao;
}

public List getUsers() {
    return dao.getUsers();
}

public User getUser(String userId) {
    return dao.getUser(Long.valueOf(userId));
}

public void saveUser(User user) {
    dao.saveUser(user);
}

public void removeUser(String userId) {
    dao.removeUser(Long.valueOf(userId));
}
}

Robot.java

public class Robot extends BaseObject {
    private static final long serialVersionUID = -1932852212232780150L;
    private Long id;
    private String name;
    private Date birthday;
    private User owner;

    public Long getId() {
        return id;
    }
    public void setId(Long id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Date getBirthday() {
        return birthday;
    }
    public void setBirthday(Date birthday) {
        this.birthday = birthday;
    }
    public User getOwner() {
        return owner;
    }
    public void setOwner(User owner) {
        this.owner = owner;
    }
}

Answer 1

我看到的唯一原因是'userManager'没有被实例化，所以这里有一个null。

您可以检查它是否已通过(userManager == null)。在我看来，您的代码很好，而不是研究逻辑。我认为您对Spring的IoC的配置就是这里的问题。

你能把你的*.xml文件。这将有助于解决你的问题。

干杯!