Reputation: 2686
i have this simple php code. Locally, with a simple xampp 1.7.3 the echo returns correctly "false" or "true"
. When i put the code online (on a server, i mean, and i do not have really knowdledge of how the server is made) it returns always "1". Why?
<?php
include "connectionToDb.php";
$nome_utente=$_GET['nome_utente'];
$queryUserAvailable = "SELECT * FROM utente where nome_utente='$nome_utente'";
$rsUserAvailable = connetti($queryUserAvailable);
if(mysql_num_rows($rsUserAvailable) == 0){
$valid=true;
}
else{
$valid=false;
}
echo json_encode($valid);
?>
ConnectionToDb.php
<?php
function connetti($SQL){
$conn = mysql_connect("localhost", "root", ""); //(online this data are obviously different)
$db = mysql_select_db("dbName",$conn);
$risultato = mysql_query($SQL,$conn)
or die("Query non valida: " . mysql_error());
return ($risultato);
}
?>
Upvotes: 1
Views: 237
Reputation: 11832
As it seems from the comments, JSON is not included in the server's PHP configuration.
You might want to consult your host and check whether you can include this by just overriding settings via a .htaccess
directive
Maybe you can also first try:
if (!extension_loaded('json')) {
if (strtoupper(substr(PHP_OS, 0, 3)) === 'WIN') {
dl('json.dll');
} else {
dl('json.so');
}
}
Upvotes: 1