这个jpql查询出了什么问题？(JPA)

Question

你能帮我找一下我的应用程序中登录方法的JPQL查询中的错误吗？

// Login
public boolean saveUserState(String email, String password) {
    // 1-Send query to database to see if that user exist
    Query query = em
            .createQuery("SELECT r FROM Role r WHERE r.email=:emailparam r.password=:passwordparam");
    query.setParameter("emailparam", email);
    query.setParameter("passwordparam", password);
    // 2-If the query returns the user(Role) object, store it somewhere in
    // the session
    Role role = (Role) query.getSingleResult();
    if (role != null && role.getEmail().equals(email)
            && role.getPassword().equals(password)) {
        FacesContext.getCurrentInstance().getExternalContext()
                .getSessionMap().put("userRole", role);
        // 3-return true if the user state was saved
        return true;
    }
    // 4-return false otherwise
    return false;
}

执行时我收到此错误:

严重:JSF1073:在处理INVOKE_APPLICATION 5期间捕获到javax.faces.event.AbortProcessingException:UIComponent-ClientId = j_idt13:j_idt17,Message =/WEB-INF/templates/BasicTemplate.xhtml @ 61,63 actionListener ="#{securityController.logIn ()}":javax.ejb.EJBException SEVERE:/WEB-INF/templates/BasicTemplate.xhtml @ 61,63 actionListener ="#{securityController.logIn()}":javax.ejb.EJBException javax.faces.event. AbortProcessingException:/WEB-INF/templates/BasicTemplate.xhtml @ 61,63 actionListener ="#{securityController.logIn()}":javax.ejb.EJBException ................ ..............由以下引起:java.lang.IllegalArgumentException:在EntityManager中创建查询时发生异常:异常描述:解析查询时出现语法错误[SELECT r FROM Role r WHERE r. email =:emailparam,r.password =:passwordparam],第1行,第46列:[,]处的语法错误.内部异常:MismatchedTokenException(79!= - 1)

Answer 1

你可能忘了添加AND或OR

喜欢:

Query query = em
            .createQuery("SELECT r FROM Role r WHERE r.email=:emailparam AND r.password=:passwordparam");