该JPA查询出了什么问题，需要通过特定名称查找用户

Question

I have a method that search a name, when the user passes a parameter. Now I need a method to tell me what's the employee's role in the company

The idea is to take this data and insert into a combobox.

This is the method that i used to find the names passes by parameter:

public List<User> findbyName(String name) { 
    EntityManager em = getEntityManager();
    String sql = "select u from User u where UPPER(u.name) like UPPER ('" + name + "%')";
    List<User> u = null;
    try {
        u = (List<User> ) em.createQuery(sql).getResultList();

    } catch (Exception e) {

        return null;          
    }
    return u;
}

And here is that method that im trying to do:

public List<User>  findbyRole(){
    EntityManager em = getEntityManager();
            String sql = "select u from User u where UPPER(u.role) like (journalist)";
            List<User>  u = null;
    try {
        u = (List<User> ) em.createQuery(sql).getResultList();

    } catch (Exception e) {

        return null;          
    }
    return u;
}

Code to insert into combobox, its right?

List <User> list= new UserJpaController().findbyRole();
     for(int i=0;i<lista.size();i++){
      combojournalist.addItem(list.get(i).getCargo());
}

Answer 1

If you want to use a literal string journalist in your query, you either need to put it into quotes, or pass it as parameter. Otherwise JPA would think that journalist is an alias, but such alias is not defined in the query.

So your query should be like this (see quotes around journalist):

select u from User u where UPPER(u.role) like ('journalist')

or if you use parameters, which is always a better option:

String sql = "select u from User u where UPPER(u.role) like (:role)";
List<User>  u = em.createQuery(sql, User.class)
         .setParameter("role", "journalist")
         .getResultList();