MySql_fetch_array() error in PHP

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:

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

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: