Re: [PHP] subscription problem

2002-06-11 Thread Miguel Cruz

On Tue, 11 Jun 2002, Doug DeVries wrote:
> Why am I getting all the messages from this newsgroup delivered to my
> mailbox?
> 
> I subscribed to the newsgroup -- or at least that is all I wanted.
>
>> To unsubscribe, visit: http://www.php.net/unsub.php

The above link would be a good start.

You don't really "subscribe" to a newsgroup, since reading them is an 
entirely passive activity.

miguel


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[PHP] subscription problem

2002-06-11 Thread Doug DeVries

Why am I getting all the messages from this newsgroup delivered to my
mailbox?

I subscribed to the newsgroup -- or at least that is all I wanted.

Thanks.

-Doug

- Original Message -
From: "Jason Soza" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Tuesday, June 11, 2002 9:36 AM
Subject: Re: [PHP] Different Problem [Re: Passing a Variable to PHP Via the
URL]


> Rather than just setting globals on in php.ini, try this:
>
> Your printf() line is:
> printf("Variables:  %s\n", $HTTP_GET_VARS["id"]);
>
> And your if() statement is:
> if($id) {}
>
> Where is $id set? It's probably not. $HTTP_GET_VARS["id"] doesn't set
> $id. If you want the "id" variable in your if(), you need:
> if($HTTP_GET_VARS["id"]) {}
>
> Or alternately do:
> $id = $HTTP_GET_VARS["id"]
> if($id) {}
>
> Although that's not necessary. Also, try using $_GET["id"], as
> $HTTP_GET_VARS[] has been deprecated in newer versions.
>
> HTH,
>
> Jason Soza
>
> - Original Message -
> From: <[EMAIL PROTECTED]>
> Date: Tuesday, June 11, 2002 8:16 am
> Subject: [PHP] Different Problem [Re: Passing a Variable to PHP Via the
> URL]
>
> > I found out that in fact PHP is creating a variable with the name
> > and value I'm passing through a URL from the querystring.  But
> > it's still not working as planned.
> >
> > The url server/test.php?id=1 creates the following results in my code:
> >
> > printf("Variables:  %s\n", $HTTP_GET_VARS["id"]);
> >
> > This line works - there IS a variable named 'id' in my page and it
> > has the correct value, 1.
> >
> > if ($id) {}
> >
> > This fails.  If I test 'id' instead of '$id' then it works but my
> > page doesn't seem to equate 'id=1' with the presence of $id.
> >
> > $result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);
> >
> > This doesn't work - again it seems $id isn't being treated
> > properly.  I get this error:
> >
> > Warning: mysql_fetch_array(): supplied argument is not a valid
> > MySQL result resource
> >
> > If I hardwire my page with the line '$id=1;' before the if
> > statement and the query everything works.
> >
> > So why isn't the variable from my URL being treated properly?
> >
> > Jesse
> >
>
>
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>



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