Global variables help...

Hi,

Quick Version:
My included function cannot read my global mysql link variable, and is
causing me lots of hassle! Any ideas?

Long Version:
I have a mysql connection set up, "$db_link" which I have declared as global
in the following way:

global $db_link;
$db_link = mysql_connect(...);

I then have a function to close the link and unset the variable so that our
"clean up" script at the end of the page will not try to close it again and
therefore error. The function reads:

function closeDB()
{
mysql_close($db_link);
unset($db_link);
}

However, the script is erroring saying it cannot find $db_link. So I tried
passing the variable from the calling statement to the function:

closeDB($db_link)
function closeDB($db_link)
{
mysql_close($db_link);
unset($db_link);
}

This closes the database link, but unsets the local $db_link in the
function, not the global variables.

I am fairly new to PHP, so if anyone can tell me why my function cannot read
the global variable, or if I am doing anything really stupid, I would really
appreciate your help!

Thanks,

Andy Hall.

"Andy Hall" <andy@jaar.co.uk> wrote in message
news:bhibv2$16j$1@sparta.btinternet.com...
> Hi,
> Long Version:
> I have a mysql connection set up, "$db_link" which I have declared as
global
> in the following way:
>
> global $db_link;
> $db_link = mysql_connect(...);
>
> I then have a function to close the link and unset the variable so that
our
> "clean up" script at the end of the page will not try to close it again
and
> therefore error. The function reads:
>
> function closeDB()
> {
> mysql_close($db_link);
> unset($db_link);
> }
>
> However, the script is erroring saying it cannot find $db_link. So I tried
> passing the variable from the calling statement to the function:
>
> closeDB($db_link)
> function closeDB($db_link)
> {
> mysql_close($db_link);
> unset($db_link);
> }
>
> This closes the database link, but unsets the local $db_link in the
> function, not the global variables.
>
> I am fairly new to PHP, so if anyone can tell me why my function cannot
read
> the global variable, or if I am doing anything really stupid, I would
really
> appreciate your help!
>
> Thanks,
>
> Andy Hall.
>
>

Maybe this will help?

function closeDB() {
global $db_link; // use the global var $db_link
mysql_close($db_link);
unset($db_link);
}

If you want to use global vars in functions declare them in the function by
"global $theVar;"

Edwin

"Andy Hall" <andy@jaar.co.uk> wrote in
news:bhibv2$16j$1@sparta.btinternet.com:

> Long Version:
> I have a mysql connection set up, "$db_link" which I have declared as
> global in the following way:
>
> global $db_link;
> $db_link = mysql_connect(...);

See the other reply on that.

> However, the script is erroring saying it cannot find $db_link. So I
> tried passing the variable from the calling statement to the function:
>
> closeDB($db_link)
> function closeDB($db_link)
> {
> mysql_close($db_link);
> unset($db_link);
> }

You're not passing the real link resource. In this case, you should be
passing it by reference (similar to pointers in C). You can do it in two
ways AFAIK:

closeDB (&$db_link);

OR

function closeDB (&$db_link) {

KAH

>
> > Long Version:
> > I have a mysql connection set up, "$db_link" which I have declared as
> > global in the following way:
> >
> > global $db_link;
> > $db_link = mysql_connect(...);
>
> See the other reply on that.
>

I tried that but it still did the same. I was thinking maybe
register_globals is set to off. I didnt set up the server and I guess if
that is the case it would not work?

Would it be better to use a _REQUEST var for the db_link as follows:

$_REQUEST["db_link"] = mysql_connect(...);

This seems to work exactly how I would like, but are there any knock on
effects/problems if I do this?

Thanks for the help

Andy.