How to pass a variable with ajax to another php file

Solved Here I leave the solution in case you come to present this same difficulty, believe me I suffered a lot until I have to analyze it in detail and use different ways to reach your solution.

Here's the select code

<div class="form-group" >
                            <label for="selectFacultad" class="col-sm-1 control-label">Facultad:</label>
                                    <div class="col-sm-4"><select name="selectFacultad" id="selectFacultad" class="form-control"  >
                                        <option value="0">Seleccione</option>
                                    </select></div>
                             <button class="btn btn-primary fa fa-search" type="submit" id="buscar" onclick="Cargar();"></button>
                        </div>

Both the name of the select and the id must be called equal only to avoid generating confusion, likewise the script goes like this:

<script>
        function Cargar(){
            var url="llenarPrograma.php";
            var selectFacultad= document.getElementById("selectFacultad").value;

            $.ajax({
                type: "POST",
                url:url,
                data:{selectFacultad:selectFacultad},
                success: function(datos){
                    $('#contenido').html(datos);
                }
            });
        }
     </script>

And finally the file completaPrograma.php

<?php
 include_once('conexion.php');

$codigo=$_POST["selectFacultad"];

$resultado=$conexion->query("SELECT id_programa,nombre_programa,duracion_semestral,horario,metodologia FROM programa WHERE id_facultad='$codigo'");

$total=$resultado->num_rows;

if($total>0){
$tabla="<table class='table table-bordered table-success nowrap'>";
$tabla.="<tr><td>Codigo</td><td>Nombre</td><td>Duración</td><td>Horario</td><td>Metodologia</td></tr>";
while($row=$resultado->fetch_assoc()){
    $tabla.="<tr>" ;
    $tabla.="<td>".$row['id_programa']."</td>";
    $tabla.="<td>".$row['nombre_programa']."</td>";
    $tabla.="<td>".$row['duracion_semestral']."</td>";
    $tabla.="<td>".$row['horario']."</td>";
    $tabla.="<td>".$row['metodologia']."</td>";
    $tabla.="</tr>" ;
}
    $tabla.="</table>" ;
    echo $tabla;
}
else{
     echo "<p>Aún no hay registros en la base de datos</p>";
}
?>

I hope and serve you all, thanks to the people who responded to me, it was very helpful

When you perform the ajax in the "data" only the value of the variable codigoFacultad is sent, a key " codigoFacultad " is not sent that in php you try to access $_POST['codigoFacultad']

try placing the following in the data: {"codigoFacultad" :codigoFacultad }

I hope this code will help you:

to your variable codeFaculty modify it by the following:

var codigoFacultad = { 
   "codigoFacultad":$('#selectFacultad').val();
};

you will be sending the name that identifies the variable (which the POST recognizes) and the value of that variable.

You have to declare where you are going to send the post

<script type="text/javascript">

    function Cargar(){
         var codigoFacultad = $('#selectFacultad').val();
        alert(codigoFacultad);

        /*Le indicas el nombre del del post*/
        var form_data = new FormData();
        form_data.append('codigoFacultad', codigoFacultad);

        $.ajax({
            type: "POST",
            dataType: "html",
            url: "llenarPrograma.php",
            /*Envias los datos*/
            data: form_data,
            success: function(data){
                 $('#contenido').load('llenarPrograma.php');
            }
        });

        }
</script>

In php you receive it like this.

$cod_facultad = $_POST['codigoFacultad'];

And in your HTML code you lack the value in the select option, they should be like this.

<select name="codigoFacultad" id="selectFacultad" class="form-control"  >
   <option value="0">Seleccione</option>
   <option value="1">Facultad 1</option>
   <option value="2">Facultad 2</option>
</select>

I recommend that your name and id be the same, since it can lead to confusion. And also your SQL statement is very susceptible to sql injection I recommend that you review how to go through parameters or PDO.