JPA - NamedQuery 名称：未找到 User.checkCredentials

编程问答 2022-06-02

问题描述

我正在尝试使用 JPA 对用户进行凭据检查。 每次运行代码，我得到：

java.lang.IllegalArgumentException: NamedQuery of name: 未找到 User.checkCredentials。

我已经做了很多尝试，但我找不到如何找出导致错误的原因。

（我在 MysqL 上创建了数据库，我称之为“gamified_marketing”）

index.html

<!DOCTYPE html>
<html xmlns:th="http://www.thymeleaf.org">
<head>
    <Meta charset="UTF-8">
    <link rel="stylesheet" href="mystyle.css">
    <title>Login page</title>
</head>

<body>
    <h1>Welcome back to the Gamified Marketing application</h1>
    <form action="CheckLogin" method="POST">
        Username: <input type="text" name="usr" required> <br>
        Password: <input type="password" name="pwd" required> <br>
        <input type="submit" value="login">
        <p th:text=" ${errorMsg}"></p>
    </form>
</body>
</html>

检查登录

@WebServlet("/CheckLogin")
public class CheckLogin extends HttpServlet {
    private static final long serialVersionUID = 1L;
       
    @EJB(name = "it.polimi.db2.services/UserService")
    private UserService usrService;
    
    public CheckLogin() {
        super();
    }

    protected void doPost(HttpServletRequest request,HttpServletResponse response)
            throws servletexception,IOException {
        String usr = null;
        String pwd = null;
        
        try {
            usr = request.getParameter("usr");
            pwd = request.getParameter("pwd");
            if (usr == null || pwd == null || usr.isEmpty() || pwd.isEmpty()) {
                throw new Exception("Missing or empty credential value");
            }

        } catch (Exception e) {
            response.sendError(HttpServletResponse.SC_BAD_REQUEST,"Missing credential value");
            return;
        }
        
        User user;
        try {
            user = usrService.checkCredentials(usr,pwd);
        } catch (CredentialsException | NonUniqueResultException e) {
            response.sendError(HttpServletResponse.SC_BAD_REQUEST,"Could not check credentials");
            return;
        }

用户服务

@Stateless
public class UserService {
    @PersistenceContext(unitName = "GamifiedMarketingApplicationEJB")
    private EntityManager em;

    public UserService() {}

    public User checkCredentials(String usr,String pwd) throws CredentialsException,NonUniqueResultException {
        List<User> uList = null;
        
        try {
            uList = em.createNamedQuery("User.checkCredentials",User.class).setParameter(1,usr).setParameter(2,pwd).getResultList();
        } catch (PersistenceException e) {
            throw new CredentialsException("Could not verify credentals");
        }
        
        if (uList.isEmpty())
            return null;
        else if (uList.size() == 1)
            return uList.get(0);
        throw new NonUniqueResultException("More than one user registered with same credentials");
    }
}

用户

@Entity
@Table(name = "Userr",schema = "gamified_marketing")
@NamedQuery(name = "User.checkCredentials",query = "SELECT r FROM User r  WHERE r.username = ?1 and r.password = ?2")

public class User implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private String username;
    
    private String email;
    private String password;
    private String lastLogin;
    ...

persistence.xml

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.2" 
    xmlns="http://xmlns.jcp.org/xml/ns/persistence" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_2.xsd">
    <persistence-unit name="GamifiedMarketingApplicationEJB"
        transaction-type="JTA">
        <jta-data-source>GamifiedMarketing</jta-data-source>
        <class>it.polimi.db2.entities.User</class>
        <properties>
            <property name="eclipselink.logging.level" value="FINE" />
        </properties>
    </persistence-unit>
</persistence>

解决方法

暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！

如果你已经找到好的解决方法，欢迎将解决方案带上本链接一起发送给小编。

小编邮箱:dio#foxmail.com (将#修改为@）

javajavajava-ee-6jpajpa