Query resources and session variables

[wdseelig]

Can anybody tell me why it does not seem to be possible to store the result returned by mysql_query in a session variable and then reuse it?

Suppose that I declare $dbResult to be a session variable, and then, on the first entry to the page I initialize it with a statement like this:

$dbResult = mysql_query($query,$link)

When I subsequently re-enter the page and attempt to use $dbResult, I get an error message that says:

Warning: Supplied argument is not a valid MySQL result resource

When I print out $dbResult on the first entry to the page, I get a value of
Resource id #2. However, when I re-enter the page, I get a value of 0!?!

Here is sample code illustrating the problem:

<?
if ( ! (isset($linkVar) ) )
{

	session_start();
	if ( ! session_is_registered("linkVar")  ) {
		session_register("LinkVar");
		session_register("dbResult");
	}

/
Make the mySQL connection ...
/
if (! ( $dbLink = mysql_pconnect("localhost","xxx","yyy") ) ) {
print("Failed to connect to the mySQL server!<BR>\n");
print("Abandoning connection attempt");
exit();
}
/
... and select the GOP Data Base
/
if ( ! mysql_select_db("dbase",$dbLink) ) {
print("Unable to use data base");
print("Abandoning attempt to use the data base!<BR>\n");
exit();
}

// Now store the $dbLink in a session variable

$linkVar = $dbLink;
}						


if ( $submitted < 1)
{
	$query="Select * from personal_info_table";
	if ( ! $dbResult = mysql_query($query,$linkVar)  )
	{
		print("Unable to query the People Table in the data base<br>\n");
		print("MySQL reports the following error:  " . mysql_error() . "<br>\n");
		exit();
	}
	$row = mysql_fetch_array($dbResult,MYSQL_ASSOC);
}
else
{
	print("I am in the second part of the routine");
}

?>

<form method="get" action=<? print($PHP_SELF) ?> >
<INPUT type="hidden" name="submitted" value="1" >

<?
print("The value of the dbResult variable is: " . $dbResult . "<br>\n");
$row = mysql_fetch_array($dbResult,MYSQL_ASSOC);
print("The value of the last name is: " . $row["last_name"] );
?>
<BR><BR>
<INPUT type="submit" value="Submit the Form">
</form>

[Anon]

The reason is because resources are handled differently. Let's say you have:

$result = mysql_query("something");

So $result is a resource, right? But in reality, it is a resource pointer. It is not the actual resource, but it is a pointer to the memory address of the resource.

When you save that resource pointer in a cookie, you're just saving the address of the resource. When the script finishes, the resource is deleted so the next time the script runs you'll get a "Invalid Resource" error.

Diego

[wdseelig]

Thank you very much.