Your code has a lot of errors that I will enumerate starting with this block:
$.ajax({
var a = document.write(dias[day]);
url:"includes/apen_files/recarga_agenda_pordias.php",
data:{envio:a},
type:"post",
success:function(a){
document.getElementById("demo").value = a;
}
});
The parameter passed to .ajax() is a class that is defined between keys { datos } that could erroneously induce you to be similar to the definition of a function, so that the value assignment you make ( var a = document.write(dias[day]); ) It is incorrect and will give you an error.
Also, you are assigning the variable a what returns document.write and not the day of the week obtained.
To continue, success from jQuery 3.0 is called done :
Deprecation Notice: The jqXHR.success() , jqXHR.error() , and jqXHR.complete() callbacks are removed as of jQuery 3.0. You can use jqXHR.done() , jqXHR.fail() , and jqXHR.always() instead.
In Spanish:
Obsolescence notice: The jqXHR.success() , jqXHR.error() and jqXHR.complete() methods have been removed in jQuery 3.0. You can use jqXHR.done() , jqXHR.fail() and jqXHR.always() instead.
It can be used as a method, so your code would be:
var a = dias[day];
$.ajax({
url: "includes/apen_files/recarga_agenda_pordias.php",
data: {envio:a},
type: "post",
}).done(function(a) {
document.getElementById("demo").value = a;
});
On the other hand, I have to tell you that the only sure way to send PHP data to javascript is by using json_encode() and never by printing the content of a variable directly:
var mes = <?= json_encode($mes) ?>;
var dia = <?= json_encode($dia) ?>;
var ano = <?= json_encode($any) ?>;
It is not necessary to put the values in quotes because json_encode() will put quotation marks if the content is a string of characters. In case of Boolean or numeric values it will not put them (they are not necessary).
Finally, it is not recommended to call the Date() constructor with a localized date string. It is advisable to use the constructor in which the day, month and year are used as parameters:
var day = new Date(
<?= json_encode($any) ?>,
<?= json_encode($mes) ?>,
<?= json_encode($dia) ?>
).getDay();
This is the code that I used to reproduce your problem and fix it:
nuevo.php
<?php
/* El 28 de febrero de 2018 fue miércoles */
$mes = 02;
$dia = 28;
$any = 2018
?><!DOCTYPE html>
<html lang="es">
<head>
<title>Título de ejemplo</title>
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no" />
<script src="https://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8=" crossorigin="anonymous"></script>
</head>
<body>
<script>
var dias = ["Domingo", "Lunes", "Martes", "Miércoles", "Jueves", "Viernes", "Sábado"];
var day = new Date(
<?= json_encode($any) ?>,
<?= json_encode($mes) ?>,
<?= json_encode($dia) ?>
).getDay();
console.log('Enviando el valor', dias[day]);
$.ajax({
url: 'receptor.php',
data: {
envio: dias[day],
},
type: 'post',
dataType: 'json',
}).done(function(a) {
console.log(a);
document.getElementById("demo").value = a;
});
</script>
<input type="text" id="demo" value="Valor inicial" size="40" />
</body>
</html>
receptor.php
<?php
/* Indicamos que el valor devuelto será JSON codificado en UTF-8 */
header('Content-type:application/json; charset=utf-8');
/* Obtenemos el valor enviado por XHR */
if (isset($_POST['envio'])) {
$variable = $_POST['envio'];
} else {
$variable = '(no se recibieron datos por POST)';
}
/* Devolvemos una cadena montada con dicho valor */
echo json_encode("Hemos recibido: $variable");