On 16 Nov 2001 [EMAIL PROTECTED] wrote:
> But, can I free a result set that returned mysql_query in mysql_result_free?
>
> Does this code works? :
> // here I'm successfully conected using mysql_pconnect();
> $result = mysql_query("insert into table1(col1) values(1)");
> if($result) {
> // here I get a warning only if I used a insert statment
> mysql_free_result($result);
> }
You only can free a result returned by a SELECT statement (as is mentioned
in the manual).
Derick
>
>
>
> Previous Comments:
> ------------------------------------------------------------------------
>
> [2001-11-16 13:11:03] [EMAIL PROTECTED]
>
> >From the manual: (http://uk.php.net/manual/en/function.mysql-query.php):
>
> mysql_query() returns TRUE (non-zero) or FALSE to indicate whether or not the
>query succeeded. A return value of TRUE means that the query was legal and could be
>executed by the server. It does not indicate anything about the number of rows
>affected or returned. It is perfectly possible for a query to succeed but affect no
>rows or return no rows.
>
> <snip>
>
> For SELECT statements, mysql_query() returns a new result identifier that you can
>pass to mysql_result(). When you are done with the result set, you can free the
>resources associated with it by calling mysql_free_result().
>
> Not a bug, but intended behavior >> bogus
>
> ------------------------------------------------------------------------
>
> [2001-11-16 11:30:05] [EMAIL PROTECTED]
>
> Tested in RedHat 7.1 with official updates.
>
> ------------------------------------------------------------------------
>
> [2001-11-16 11:29:01] [EMAIL PROTECTED]
>
> Summary changed
>
> ------------------------------------------------------------------------
>
> [2001-11-16 11:25:43] [EMAIL PROTECTED]
>
> If you use the result set of a insert query, you get a warning message.
>
> // connect
> // do an insert
> $result = mysql_query("insert into table1(col1) values(1)");
> // free a result
> mysql_free_result($result);
> // a warning is generated
>
> Warning: Supplied argument is not a valid MySQL result resource in your_source.php
>on line line_number
>
> php-mysql-4.0.4pl1-9
> mysql-3.23.36-1
> RedHat 7.1
>
>
> './configure' '--prefix=/usr' '--with-config-file-path=/etc' '--disable-debug'
>'--enable-pic' '--enable-shared' '--enable-inline-optimization'
>'--with-apxs=/usr/sbin/apxs' '--with-exec-dir=/usr/bin' '--with-regex=system'
>'--with-gettext' '--with-gd' '--with-jpeg-dir=/usr' '--with-png' '--with-zlib'
>'--with-db2' '--with-db3' '--with-gdbm' '--enable-debugger' '--enable-magic-quotes'
>'--enable-safe-mode' '--enable-sockets' '--enable-sysvsem' '--enable-sysvshm'
>'--enable-track-vars' '--enable-yp' '--enable-ftp' '--enable-wddx' '--without-mysql'
>'--without-oracle' '--without-oci8' '--with-xml'
>
>
> ------------------------------------------------------------------------
>
>
>
> Edit this bug report at http://bugs.php.net/?id=14086&edit=1
>
>
> --
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