Question

错误:(在我的spring crud操作项目中抛出异常导致根本原因错误）

Error: Nov 14, 2017 12:43:12 PM org.apache.catalina.core.StandardWrapperValve invoke

SEVERE: Servlet.service() for servlet [spring] in context with path 


[/Spring4MVCAngularJSExampleLatest] threw exception [Request processing failed; nested exception is java.lang.NullPointerException] with root cause
java.lang.NullPointerException
    at com.websystique.springmvc.service.UserServiceImpl.saveUser(UserServiceImpl.java:56)
    at 
com.websystique.springmvc.controller.HelloWorldRestController.createUser(HelloWorldRestController.java:69)

（这笔款项是在订婚结束时一次性支付的。我亲切的主人，应我的要求，给了我一笔一笔款项来代替我的年金。我建议一次性付款，条件是你退出这个诉讼不是一次性拨款，而是通过拨款或资金。）

代码：

JavaService Impl:


public void saveUser(User user) 
{


System.out.println(user.getAddress()+""+user.getId());//printing successfully

        // user.setId(1);
        ((UserDaoImpl) users).saveOrUpdate(user);// i got error here
         //ud.saveOrUpdate(user);   
    }

=============================================== ==

UserService.java这是服务类

public interface UserService {

    User findById(long id);

    User findByName(String name);

    void saveUser(User user);

    void updateUser(User user);

    void deleteUserById(long id);

    List<User> findAllUsers(); 

    void deleteAllUsers();

    public boolean isUserExist(User user);

}

=============================================== =========

UserDaoImpl.java :(这是存储库impl类）

public List<User> list() {
        String sql = "SELECT * FROM user12";
        List<User> listContact = jdbcTemplate.query(sql, new RowMapper<User>() {

            @Override
            public User mapRow(ResultSet rs, int rowNum) throws SQLException {
                User aContact = new User();

                aContact.setId(rs.getInt("id"));
                aContact.setUsername(rs.getString("username"));
                aContact.setEmail(rs.getString("email"));
                aContact.setAddress(rs.getString("address"));

                System.out.println("ASADADADAS" + rs.getString("username"));
                return aContact;
            }

        });

=================================

HelloWorldRestController.java（这是控制器类）

  @RequestMapping(value = "/user/", method = RequestMethod.GET)

    public ResponseEntity<List<User>> listAllUsers() {
        List<User> users = userService.findAllUsers();
       /* if(users.isEmpty()){
            return new ResponseEntity<List<User>>(HttpStatus.NO_CONTENT);//You many decide to return HttpStatus.NOT_FOUND
        }*/
        return new ResponseEntity<List<User>>(users, HttpStatus.OK);
    }

=============================

UserDao.java（这是存储库类）

public interface UserDao {

    public void saveOrUpdate(User user);

    public void delete(int id);

    public User get(long id);

    public List<User> list();

}

Answer 1

In the service the ((UserDaoImpl) users) is null. It means it was not injected properly.

The UserDaoImpl must be annotated with @Repository (or @Component) to be a spring bean. Also it should be in component scan package to let Spring find it and create a bean.

在我的spring crud操作项目

1 个答案: