[英]Spring <form:select> tag causing error in Spring mvc + Hibernate app
[英]Spring MVC + Hibernate App
我也有一个应用程序(Spring MVC + Hibernate)和AppFuse Framework。
我有两个实体:具有多对一关系的User
和Robot
。
我需要为Robot
表单(robotForm.jsp)的所有者( User
)添加一个下拉列表。
Robot
实体具有一个User
。 我读到我必须为User
创建一个自定义Editor
。 ( UserCustomEditor extends PropertyEditorSupport
),并覆盖referenceData
在RobotFormController
添加在initBinder
了。
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referenceData
方法的ownersMap.put("owners", userManager.getUsers();
行中得到了NullPointerException
。
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));
}
}
机器人.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;
}
}
我看到的唯一的原因是您没有实例化'userManager'
。
您可以检查它是否已经通过(userManager == null)
。 您的代码对我来说看起来不错,没有考虑逻辑。 我认为您的Spring IoC配置是这里的问题。
您可以发布* .xml文件吗? 这将有助于解决您的问题。
干杯!
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