24.06.2009, 08:00
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Участник форума
Регистрация: 28.01.2008
Сообщений: 247
Провел на форуме:
205760
Репутация:
28
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PHP код:
function fetch_user_salt($length = 3)
{
$salt = '';
for ($i = 0; $i < $length; $i++)
{
$salt .= chr(rand(32, 126));
}
return $salt;
}
$nhd1 = mysql_connect($host_a, $user_db_a, $pass_db_a);
$nhd2 = mysql_connect($host, $user_db, $pass_db);
mysql_select_db("$dbase", $nhd1);
mysql_select_db("$dbase", $nhd2);
//do a query from db1:
$sql = "SELECT id, email, username, pwd, regdate FROM $dtable"; $which = $nhd1;
mysql_query($query,$which);
$result = mysql_query($sql);
while(list($id, $email, $username, $pwd, $regdate) = mysql_fetch_row($result)) {
$link = mysql_pconnect ($host, $user_db, $pass_db);
if ($link) {
mysql_select_db($dbase);
} else {
exit("Unable to connect to database. Please try again later.\n");
}
if (is_array($rows))
{
foreach ($rows as $string)
{
$salt = fetch_user_salt( 3 );
$pwd = md5( md5( trim($pwd) ).$salt );
//do a query from db2
$query = "INSERT INTO f_user SET userid='$id',email='$email',username='$username',password='$pwd',salt='$salt'"; $which = $hnd2;
mysql_query($query,$which);
}
}
}
Пишет на ошибку
Parse error: syntax error, unexpected ')', expecting '=' in Z:\home\localhost\www\myimport\index.php on line 39
39 while(list($id, $email, $username, $pwd, $regdate) = mysql_fetch_row($result)) { |
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