saya jalanin di kompi saya pakai:
- PHP 4.0
- mysql-3.23.38
hasilnya OK
di web hosting
- PHP version 4.4.2
- MySQL version 4.1.14-standard
hasilnya error
login and passwod sudah saya cek berX2, sudah betul
error nya disini:
headfoot.php
<?php
function get_main_content() {
$mncontent = mysql_result(mysql_query("select content from html where
type='1page'"), 0); // line 3
echo($mncontent);
}
function uheader() {
global $m_header, $ref;
include($m_header);
}
function ufooter() {
global $m_footer, $ref;
include($m_footer);
}
-----
apa mungkin beda Version di PHP/MYSQL, beda script/string juga?
sebagai tambahan Index.php
<?php
session_start();
session_register("sess_data");
session_register("ref");
session_register("sess_name");
session_register("sess_passwd");
include("vars.php");
include("headfoot.php");
mysql_connect($db_host, $db_user, $db_pwd);
mysql_select_db($db_name);
if ($_SERVER['REMOTE_ADDR']=='127.0.0.1')
{$_SERVER['REMOTE_ADDR']=$_SERVER['HTTP_X_FORWARDED_FOR'];}
----
Terima KAsih
----- Original Message -----
From: "Oebay" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Wednesday, March 29, 2006 8:26 AM
Subject: [tanya-jawab] Re: Mysql error
> Warning: mysql_result(): supplied argument is not a valid MySQL result
> resource in /home/betanea/public_html/headfoot.php on line 3
>
> dan
>
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
> resource in /home/betanea/public_html/headfoot.php on line 42
>
> kira2 apa ya solusinya?
Yup, sepertinya error di php scriptnya, silakan di check dulu scriptnya.
atau mungkin querynya bisa di print dulu ke layar biar bisa lebih
jelas errornya.
<?php
$query = "select * from anu";
print $query."<br>";
?>
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