在春季将多个输入绑定到列表

Question

因此，我有两个部门department和department，一个department可以为此具有多个名称，我制作了此表格，您可以在其中添加更多名称的输入，我查找了如何将其绑定到模型，但显然不建议这样做。

https://i.stack.imgur.com/AgO1F.png

这些是部门和指定类别

function users($db, $id, $itemsa) {

    $sql = "select *  from users where id = :aid limit 1";
    $st = $db->prepare($sql);
    $st -> execute([":aid" => $id]);

    // you only have one row returned, so a simple fetch
    $row = $st->fetch(PDO::FETCH_ASSOC);

    $arra = explode(', ', $itemsa);
    $items = "";

    foreach ($itemsa as $col )  {
        $items .=
        "<div class='rtitle' data-id = " . $row['id'] . ">" . mb_strtoupper($row[$col]) . "</div>\n" .
        "<div class='rstory' data-id = " . $row['id'] . ">" . $row[$col] . "</div>\n";

    }
    echo $items;
}

$itemsa = "date, name, phone, address, color, height";

users($db, $id, $itemsa);

}

public class Departement implements Serializable{
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id_departement")
private Long idDepartement;
@Column(name="nom_dept")
private String nomDept;


@Column(name="designation")
@OneToMany(fetch = FetchType.EAGER, mappedBy = "departement")
private List<Designation> designation;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "departement")
private List<EmployeSociete> employees;

public Departement() {
}
public Departement(String nomDept, List<Designation> designation, List<EmployeSociete> employees) {
    super();
    this.nomDept = nomDept;
    this.designation = designation;
    this.employees = employees;
}

public Long getIdDepartement() {
    return idDepartement;
}

public void setIdDepartement(Long idDepartement) {
    this.idDepartement = idDepartement;
}

public String getNomDept() {
    return nomDept;
}

public void setNomDept(String nomDept) {
    this.nomDept = nomDept;
}

public List<Designation> getDesignation() {
    return designation;
}

public void setDesignation(List<Designation> designation) {
    this.designation = designation;
}

public List<EmployeSociete> getEmployees() {
    return employees;
}

public void setEmployees(List<EmployeSociete> employees) {
    this.employees = employees;
}

}