I can't seem to figure out what's the problem. For my next little project I'm creating dynamic web page with database and etc. I need to get all the necessary variables from PHP file. But for some reason I cannot do it if I include another PHP file. (I need it for database queries).
main.php
include ('databaseQueries.php');
if (isset($_POST["points"])){
$points = json_decode($_POST["points"]);
if($_POST["execute"] == 1){
}
}
$advert= array(
'Hello' => 'Hello world!',
'bye' => 'Why bye?',
);
echo json_encode($advert, $another);
pageJs.js
$.ajax({
url : 'php/main.php',
type : 'POST',
dataType : 'json',
success : function (result) {
console.log(result);
},
error : function (err) {
console.log("Failed");
}
})
databaseQueries.php
$another = "another string";
If I remove the include and $another variable from json_encode. Everything works and I get object in console log. But if I leave the those two things, Ajax call fails.
What I'm doing wrong and how can I get both the $test array and $another variable?
Thank's in advance!
$advertdefined?include ('databaseQueries.php');? Are youecho'ing stuff, because if you do, the response you're sending back is not valid JSON