Your query on the previous line is failing for one reason or another.

You can use mysql_error() to get the error message from MySQL, or you
can place an echo statement in your code to make sure the SQL
statement is what you think it is.

You can also try the SQL statement yourself interactively at a MySQL


--- empty <[EMAIL PROTECTED]> wrote:

> I have a problem (or  mistake(s));
> for this code
>  39|   $result=mysql_query("slecet * from site_members where
>        User_Name='$username'");
>  40|   $num_row=mysql_num_rows($result);
>  41|   if($num_row>0) echo("error 46");
> as a reult I have unsolved error in my hands :(
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> result resource in D:\sites\inc\ on line 40

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