Have you checked to make sure that table exists in your database? And
that it's not corrupted?
You might want to issue this command instead of the "Select NOW()";
below:
"SHOW TABLES";
Make sure you are seeing the table.
On Mar 8, 2005, at 8:36 AM, Anuradika wrote:
This first code is working well, which return the date of the
server, but, getting rows from the table tbl_BBY doesnt work, please
help
$query = "Select NOW() ";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
echo $row[0];
}
exit;
$query = "Select txt_ID FROM tbl_BBY ";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
echo $row['txt_ID'];
}
--- In php-objects@xxxxxxxxxxxxxxx, Anuradika Jaganadhan
<anuradika80@xxxx> wrote:
> i have tried to remove the @ which looks like this now
>
> $cx = mysql_connect('localhost', 'db', 'password');
> mysql_select_db("db") or die( "Unable to select
> database");
>
> and also used without MYSQL_ASSOC, but still im
> getting the same error message. Can anyone help me
> please
>
>
> --- Ryan Sexton <rsexton@xxxx> wrote:
>> Are you sure the Conn_to_Mysql.php contains the
>> correct username and
>> password and the correct host to where your mysql
>> installation is? I'd
>> say you're failing to make a connection to your
>> mysql database at the
>> office because it's setup differently. I'd bet you
>> have a @ in front of
>> your mysql_connect() statement in that include file.
>> remove it and run
>> your script to see what the error message is.
>>
>> Good luck,
>> Ryan
>>
>> On Mar 7, 2005, at 6:42 PM, Anuradika wrote:
>>
>>
>>
>> Hi there,
>> I have coded the project using php 5.0.3, with MySQL
>> 4.1. I am using
>>
>> require("Conn_to_Mysql.php");
>> $query = "Select * from tbl_BBY ";
>> $result = mysql_query($query);
>> while($row = mysql_fetch_array($result,
>> MYSQL_ASSOC))
>> {
>>
>> }
>>
>> Its running in my office pc, and my home pc, but
>> when i upload to the
>> webserver, its giving error as such, anyone able to
>> help me to solve
>> this issue.
>>
>> Warning: mysql_fetch_array(): supplied argument is
>> not a valid MySQL
>> result resource
>>
>> Thanks in advance.
>>
>>
>
>
PHP Data object relational mapping generator - http://www.meta-language.net/
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