07-08-2011, 14:13
Ma on się łączyć z bazą danych pobierać hasło i e-mail osoby i wyświetlić w postaci tablicy:
<?php
function db_connect() {
$conn = mysql_connect ('localhost', 'user1', 'haslo');
if (!$conn)
{
echo mysql_error ();
exit;
}
if (!$db = mysql_select_db('baza'))
{
echo mysql_error ();
exit;
}
return $conn;
}
function get_user($userEmail,$userPassword)
{
db_connect ();
$query = sprintf("SELECT * FROM users WHERE e-mail='$userEmail' AND password='$userPassword'"
);
$result = mysql_query ($query);
$row = mysql_fetch_assoc($result); // w tej linijce mi wywala błąd
return $row;
}
$a = get_user('[email protected]','osoba');
print_r ($a);
echo "<br>";
echo $a['name'];
?>
<?php
function db_connect() {
$conn = mysql_connect ('localhost', 'user1', 'haslo');
if (!$conn)
{
echo mysql_error ();
exit;
}
if (!$db = mysql_select_db('baza'))
{
echo mysql_error ();
exit;
}
return $conn;
}
function get_user($userEmail,$userPassword)
{
db_connect ();
$query = sprintf("SELECT * FROM users WHERE e-mail='$userEmail' AND password='$userPassword'"
);
$result = mysql_query ($query);
$row = mysql_fetch_assoc($result); // w tej linijce mi wywala błąd
return $row;
}
$a = get_user('[email protected]','osoba');
print_r ($a);
echo "<br>";
echo $a['name'];
?>