RE: Php error with MySql

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[snip]
I get the following error 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/wilmail/public_html/elblog.php on line 7
&n=&   //error ends here

with the following bit of code

$qResult = mysql_query ("SELECT * FROM blog_entries ORDER BY id DESC");

$nRows = mysql_num_rows($qResult);
$rString ="&n=".$nRows;

If I am just naming a variable how is the argument not valid?
[/snip]

Perhaps your query failed, but you've got no error checking.

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