MySql_fetch_array() error in PHP

[debajyoti2040]

In config.php line 28 I have written the following lines.
$query = "SELECT * FROM `users` WHERE username = '".$uname."' AND password = '".$pword."'";
# set a query

$online = mysql_fetch_array(mysql_query($query));

Using MYSQL and PHP in a remote linux server I am getting the following error.
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/india/public_html/xxxxx/auth/config.php on line 28

Can anybody advice , why I am getting the error.

[ronverdonk]

It most likely means you have an error in the mysql_select because no object was created. Statement of warning: do never nest mysql queries, like you do here, because you can't get to the error that caused it. Instead display the errors and continue depending on the result, like in the following sample:[php]$query = "SELECT * FROM `users` WHERE username = '$uname' AND password = '$pword';
$res = mysql_query($query)
or die("Error in select: " . mysql_error());
if (mysql_num_rows > 0)
$online = mysql_fetch_array($res);
else
echo 'No records available';[/php]
Ronald :cool:

[Nert]

hi Sir Ronald,

i just edited something on your post at the line "if(mysql_num_rows > 0)". (^_^)

[php]
$query = "SELECT * FROM `users` WHERE username = '$uname' AND password = '$pword';
$res = mysql_query($query)
or die("Error in select: " . mysql_error());
if (mysql_num_rows($res) > 0)
$online = mysql_fetch_array($res);
else
echo 'No records available';[/php]


--nert

[ronverdonk]

hi Sir Ronald,

i just edited something on your post at the line "if(mysql_num_rows > 0)". (^_^)

[php]
$query = "SELECT * FROM `users` WHERE username = '$uname' AND password = '$pword';
$res = mysql_query($query)
or die("Error in select: " . mysql_error());
if (mysql_num_rows($res) > 0)
$online = mysql_fetch_array($res);
else
echo 'No records available';[/php]


--nert

You are right, thanks.

Ronald :cool: