从带有 php 没有 jquery 的 select2 中获取值

Question

I have a registration system where there are several fields. One of the registration fields is Segments and they come from the database:

public function comboSegmentos($key)
{
    $sqlMostrar = mysqli_query($this->conexao, "SELECT * FROM loja_segmentos;");
    if (mysqli_num_rows($sqlMostrar) == 0) 
    {
        $mostrar = "<div style='color: red'>Nenhuma segmento cadastrado até o momento.</div>";
    } 
        else 
    {
        $mostrar = "<select name='Segments[]' id='segmentos' class='form-control select2' multiple='multiple' style='width: 35%'>";
        while ($jmMostrar = mysqli_fetch_object($sqlMostrar)) 
        {
                $mostrar .= "<option value='".$jmMostrar->Segmentos."' ".$selected.">".$jmMostrar->Segmentos."</option>";
        }
        $mostrar .= "</select>";
    }
    return $mostrar;
}

So that the method doesn't have too many parameters, I did it like this:

<?php
...
$dados = array_filter($_POST);
echo $metodos->cadastrarProdutos($dados);

Method cadastrarProdutos($dados)

But I'm not able to get the values of the Segments even using foreach().

<?php
public function cadastrarProdutos(array $dados)
{
  ...
  $segments = $dados["Segments"];
  foreach($segments as $seg)
  {
     $segm = $seg;
  }
  ...
}

How can I get the values of the select2 field and get them using a PHP method without Jquery?

Answer 1

I managed to solve it as follows:

<?php
public function cadastrarProdutos(array $dados)
{
  ...
  $segments = $dados["Segments"];
  $segm = array();
  foreach($segments as $seg)
  {
     $segm[] = $seg;
  }
  $segments = implode(',',$segm);
  ...
}