Re: [PHP] Conditions within a function
On Tuesday, April 16, 2002, at 11:32 AM, Scott St. John wrote:
> What I am also discovering is I have to use die; to halt execution for
> certain things. An example, I check to see if a user is still active in
> the system, if they are not I use a header to redirect them to a page
> that
> tells them they are inactive.
>
> if ($userActive == 0){
> header ('Location: notice.php');
> }
>
> As is it does not work until I do this:
>
> if ($userActive == 0){
> header ('Location: notice.php');
> die;
> }
>
That's strange, I never have to use "die". Try using an absolute path
in header(), like header('Location: http://domain.com/notice.php'), that
may fix this (not sure). Also make sure that header() is executed
before any HTML is output from the PHP processor, this is a common
mistake for new PHP coders. Even a single line of whitespace sent
before the header() line will screw it up, and this applies to all
header() types, not just "Location:".
Erik
Erik Price
Web Developer Temp
Media Lab, H.H. Brown
[EMAIL PROTECTED]
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Re: [PHP] Conditions within a function
On Tue, 16 Apr 2002, Erik Price wrote:
> First of all, try organizing your code using whitespace. Here is an
> example of the code, formatted to be a bit easier to read:
Thank you, that was Pine doing some nice formatting tricks of it's own.
> Now, my first question, is why do you have $userID passed by parameter
> into the function, and also globalized in the function? I am trying to
> understand this function but this seems like a mistake. You should
> probably fix this. Also, I'm not sure what the $db variable is
> globalized for -- why is it needed? It isn't used in the function, as
> far as I can see.
I see the error there and removed the global, I am moving from ASP
functions to PHP functions and still learning. I misunderstood the global
statement considering that I am passing the userID to the function.
What I am also discovering is I have to use die; to halt execution for
certain things. An example, I check to see if a user is still active in
the system, if they are not I use a header to redirect them to a page that
tells them they are inactive.
if ($userActive == 0){
header ('Location: notice.php');
}
As is it does not work until I do this:
if ($userActive == 0){
header ('Location: notice.php');
die;
}
-Scott
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Re: [PHP] Conditions within a function
On Tuesday, April 16, 2002, at 08:58 AM, Scott St. John wrote:
> I have been trying to organize my code better by using functions,
First of all, try organizing your code using whitespace. Here is an
example of the code, formatted to be a bit easier to read:
function chkFirstTime($userID)
{
global $connection,$db,$userID;
$sqlChkFirstTime = "select fk_userID,
firstTime
fromfirstTime
where fk_userID = '$userID'";
$resultFirstTime = mssql_query($sqlChkFirstTime, $connection);
$rowFirstTime = mssql_fetch_array($resultFirstTime);
$firstTime = $rowFirstTime["firstTime"];
if ($firstTime == "") {
include ('includes/incFirstTime.php');
}
}
Now, my first question, is why do you have $userID passed by parameter
into the function, and also globalized in the function? I am trying to
understand this function but this seems like a mistake. You should
probably fix this. Also, I'm not sure what the $db variable is
globalized for -- why is it needed? It isn't used in the function, as
far as I can see.
> especially where I am repeating logic. The problem seems to be when I
> run
> a condition within the function, an if statement to check for results
> of a
> variable. As is the code seems to be bypasses, but if I put a die; in
> the
> function after the line that was skipped before I get the desired
> results.
Why don't you put "print($firstTime)" immediately after the $firstTime =
$rowFirstTime["firstTime"] line. It will tell you whether or not there
is actually a value there. If there is, then you know why your "if"
test is failing. If there is no variable, then you can check to see if
your includefile is working correctly.
Erik
Erik Price
Web Developer Temp
Media Lab, H.H. Brown
[EMAIL PROTECTED]
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