Re: Help w/ displaying return vals

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: <ryan_gamo@edwards.com> Para: <php-db@lists.php.net> Cc: <Greg_Ziarko@edwards.com> 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 > > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net > For additional commands, e-mail: php-db-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >