John Davidson

Message:

I'm new to PHP and I'm working on Symfony 4 project

I've made a Java project using an Enum Class called role

public class Role {

public enum RoleEnum {

Admin, SuperAdmin, Etudiant, Entreprise } private Short id;

private RoleEnum name;



public Role(Short id, RoleEnum name) {

    this.id = id;

    this.name = name ;

}

and in my user I made it like this

public abstract class User {



private Long id;

private String nom;

private String email;

private String mdp;

private String tel;

private String photo;

private Role role;

private Boolean etatCompte;

and I've made a relation between the two tables in MySQL

and I created this method in Java too

   public Role getRoleByName(RoleEnum input) {



    Statement stm;

    try {

        stm = cnx.createStatement();



        String query = "SELECT id from role Where name='" +input+"'";

        ResultSet rst = stm.executeQuery(query);



        while (rst.next()) {

            Short id = rst.getShort("id");

            return new Role(id, input);

        }

    } catch (SQLException ex) {

        Logger.getLogger(ServiceEtudiant.class.getName()).log(Level.SEVERE, null, ex);

    }

    

    return null;

}

and when I'm adding values to the table for example to add "admin" to the role I do it like this

        Role role = s.getRoleByName(Role.RoleEnum.Admin);

can anyone help me or guide how to do it the sameway in my symfony project?

---

My Answer:
"SELECT * FROM Role WHERE name = '" + input.name() + "'";__ ResultSet rs = stm.executeQuery(query);__ if (rs.next()) {_ return new Role(rs.getShort("id"), RoleEnum.valueOf(rs.getString("name")));_ } else {_ return null;_ }_ } catch (SQLException e) {_ e.printStackTrace();_ return null;_ }_}"