Use <mysql_function(...)> or die( mysql_error()) on every mysql function
or command you're performing, this will tell you what is going wrong .
HTH
Jayme.
-----Mensagem Original-----
De: <[EMAIL PROTECTED]>
Para: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Enviada em: quarta-feira, 14 de março de 2001 19:47
Assunto: [PHP-DB] Help w/ displaying return vals
> Here's my php program, I think $result never gets assigned a value,
staying
> null or whatever.
>
> $link = mysql_connect("host");
>
> mysql_select_db("dbname");
>
> $result = mysql_query ("SELECT * FROM table");
>
> $fields = mysql_num_fields ($result);
> $rows = mysql_num_rows ($result);
>
> $table = mysql_field_table ($result, $i);
>
> echo "Your '".$table."' table has ".$fields." fields and ".$rows." records
> <BR>"
>
> - - EOP - -
>
> The problem comes on the lines that reference the variable $result.
>
> I keep getting this in the browser... leading me to believe that $result
is
> null:
>
> Warning: Supplied argument is not a valid MySQL result resource
>
>
> help!
>
> Thankx0r
>
> Ryan
>
>
>
>
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