如何在CodeIgniter中从SQL中的<select>列出名称

I need to list names from my SQL db in a in my view.

  • usuarios table:

usuario

  • cultivos table:

cultivos

I want to list all the names from 'usuarios' who has a 'cultivo'

  • My PRIMARY_KEY in 'usuarios' is "id"

  • My PRIMARY_KEY in 'cultivos' is "id_cultivo" and my FOREIGN KEY in 'cultivos' is "id_usuario"

How to list all the names that have tipoUsuaro in 'usuarios' = 3?

I need to do a INNER JOIN?

I need a result like:

<select name="genero" class="form-control valid" id="genero" aria-invalid="false">
    <option value="Femenino">Esteban</option>
    <option value="Femenino">Estiven</option>
    <option value="Femenino">Lorena</option>
    <option value="Femenino">Steven</option>
</select>

1st question:

I want to list all the names from 'usuarios' who has a 'cultivo'

SELECT u.nombre FROM cultivos c INNER JOIN usuarios u ON u.id = c.id_usuario 
GROUP BY c.id_usuario

2nd question:

How to list all the names that have tipoUsuaro in 'usuarios' = 3?

SELECT nombre FROM usuarios WHERE tipoUsuario = '3'

I found a solution.

In my model:

function obtener_jefes()
{

$query = $this->db->query('SELECT * 
FROM usuarios 
INNER JOIN cultivos ON usuarios.id = =cultivos.id_usuario 
WHERE tipoUsuario = 3 
LIMIT 1;');


return $query->result();

}

In my controller.

$data_jefes['jefes'] = $this->cultivo_model->obtener_jefes();
$this->load->view('cultivo_view/agregar', $data_jefes);

In my view

<select class="form-control">
<option value="" selected="selected">Selecccionar jefe</option>
<?php foreach($jefes as $row)
{ 
echo '<option value="'.$row->tipoUsuario.'">'.$row->nombre.'</option>';
}
?>
</select>

</div>