PHP, MySQL, upit ne vraca rezultatImam MySQL bazu u kojoj postoji tabela korisnik sa ispravnim podacima. Imam .php stranicu koja radi upit.
U firebug-u posmatram slanje parametara i upit. Posalju se dobri parametri, ali mi upit ne vraca nikakav rezultat, tj. uvek mi je $output ==no
Kada upit pustim kroz SQLYog ili neki drugi program, iz baze mi se vrati podatak.
Pretpostavljam da je negde greska u .php strani.
Samo da napomenem da je rec o tutorialu.
Code:
<?php
define( "DATABASE_SERVER", "localhost" );
define( "DATABASE_USERNAME", "root" );
define( "DATABASE_PASSWORD", "" );
define( "DATABASE_NAME", "baza" );
$mysql = mysql_connect(DATABASE_SERVER, DATABASE_USERNAME, DATABASE_PASSWORD) or die(mysql_error());
mysql_select_db( DATABASE_NAME );
$username = $_POST["korisnik"];
$password = $_POST["lozinka"];
$query = "SELECT * FROM korisnik WHERE username ='$korisnik' and password='$lozinka'";
$result = mysql_fetch_array(mysql_query($query));
$output = "<loginsuccess>";
if(!$result)
{
$output .= "no";
}else{
$output .= "yes";
}
$output .= "</loginsuccess>";
print ($output);
?>
<?php
define( "DATABASE_SERVER", "localhost" );
define( "DATABASE_USERNAME", "root" );
define( "DATABASE_PASSWORD", "" );
define( "DATABASE_NAME", "baza" );
$mysql = mysql_connect(DATABASE_SERVER, DATABASE_USERNAME, DATABASE_PASSWORD) or die(mysql_error());
mysql_select_db( DATABASE_NAME );
$username = $_POST["korisnik"];
$password = $_POST["lozinka"];
$query = "SELECT * FROM korisnik WHERE username ='$korisnik' and password='$lozinka'";
$result = mysql_fetch_array(mysql_query($query));
$output = "<loginsuccess>";
if(!$result)
{
$output .= "no";
}else{
$output .= "yes";
}
$output .= "</loginsuccess>";
print ($output);
?>
[Ovu poruku je menjao Nemanja Avramović dana 27.04.2010. u 13:17 GMT+1]
