In message <[EMAIL PROTECTED]>,
Martinus Hadiwinata <[EMAIL PROTECTED]> writes
>Hi Guys,
>
>I got this warning on several page:
>*Warning*: mysql_free_result(): 25 is not a valid MySQL result resource
>
>Does anyone know how to get rid this warning message that appears on a page
>without using error_reporting() function?
>
>I have checked my syntax and they seem fine and correct. Moreover, the page
>displays the items as it should be.
As it would do, since free result would come after displaying the data.
Suggestions
1) Don't use it <G> it isn't usually necessary, unless you are
returning huge datasets, as the memory is freed when the script ends, if
you are not using persistent connections.
2) You should be feeding it with the variable which you get from your
query
$result = mysql_query("SELECT * FROM everything");
...
mysql_free_result($result)
and $result won't contain 25, which is what you get in your error
message. It should be a resource, not a number.
--
Pete Clark
Sunny Andalucia
http://hotcosta.com/Andalucia.Spain
