For some reason this section of code is causing this error
( ! ) Warning: mysql_query() expects parameter 1 to be string, resource given in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\new site\login2\login2.php on line 62
Call Stack
# Time Memory Function Location
1 0.0006 367040 {main}( ) ..\index.php:0
2 0.0015 397976 include( 'C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\new site\template.php' ) ..\index.php:6
3 0.0017 441616 page( ) ..\template.php:67
4 0.0023 466016 include( 'C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\new site\login2\login2.php' ) ..\template.php:30
5 0.0090 480072 mysql_query ( ) ..\login2.php:62
( ! ) Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\new site\login2\login2.php on line 62
Call Stack
# Time Memory Function Location
1 0.0006 367040 {main}( ) ..\index.php:0
2 0.0015 397976 include( 'C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\new site\template.php' ) ..\index.php:6
3 0.0017 441616 page( ) ..\template.php:67
4 0.0023 466016 include( 'C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\new site\login2\login2.php' ) ..\template.php:30
5
PHP Code:
$bobby = mysql_query("SELECT userid FROM online WHERE username = '$username42'")or die(mysql_error());
if( mysql_num_rows( mysql_query( $bobby ) ) != 1 )
I'm pretty sure that it's in that section
Here's the full code
PHP Code:
<?php
require "../database.php";
//if the login form is submitted
if (isset($_POST['submit'])) {
?>
<?php
// makes sure they filled it in
if(!$_POST['username'] | !$_POST['pass']) {
die('You did not fill in a required field. <a href="../login">Back</a>');
}
// checks it against the database
if (!get_magic_quotes_gpc()) {
$_POST['email'] = addslashes($_POST['email']);
}
$check = mysql_query("SELECT * FROM users WHERE username = '".$_POST['username']."'")or die(mysql_error());
//Gives error if user dosen't exist
$check2 = mysql_num_rows($check);
if ($check2 == 0) {
die('I am sorry, the information entered was incorrect. <a href="../login">Back</a>');
}
while($info = mysql_fetch_array( $check ))
{
$_POST['pass'] = stripslashes($_POST['pass']);
$info['password'] = stripslashes($info['password']);
$_POST['pass'] = md5($_POST['pass']);
//gives error if the password is wrong
if ($_POST['pass'] != $info['password']) {
die('I am sorry, the information entered was incorrect. <a href="../login">Back</a>');
}
else
{
// if login is ok then we add a cookie
$_POST['username'] = stripslashes($_POST['username']);
$hour = time() + 3600;
setcookie(ID_my_site, $_POST['username'], $hour);
setcookie(Key_my_site, $_POST['pass'], $hour);
$username42 = $info['username'];
$username44 = $info['id'];
$username43 = time();
$username49 = $username43+7200;
$username47 = date("h:i:s", $username49);
$fraig = strtotime('-1day');
$bobby = mysql_query("SELECT userid FROM online WHERE username = '$username42'")or die(mysql_error());
if( mysql_num_rows( mysql_query( $bobby ) ) != 1 )
{
$sql = mysql_query ("INSERT INTO online (id,username,userid,time,time2,ip) VALUES ('0','".$username42."','".$username44."','".$username43."','".$username47."','" . $_SERVER['REMOTE_ADDR']."')");
}
//then redirect them to the members area
// header("Location: ../echo time");
}
}
}
else
{
}
Any help would be great!
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