RE: [PHP] SESSION variable to pass login ID
-Original Message-
From: Nova [mailto:[EMAIL PROTECTED]]
Sent: 12 January 2003 20:04
This statement doesnt look right to me.
if(isset($_POST['userid']) isset($_POST['pword'])){
$_SESSION['user'] = $_POST['userid'];
$_SESSION['password'] = $_POST['pword'];
}
Nowt wong wi' that!
the if should be:
if ((statement)(statement))
That's not true! The syntax of if is:
if (expression) statement;
and one possible expression is:
expression expression
so this is a valid if condition:
if (expression expression)
Now, you *may* need additional parentheses around the expressions if they
contain lower-priority operators than , but they are not required.
Also: a statement is *not* an expression, so the if condition can never
include statements!
Cheers!
Mike
-
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
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[PHP] SESSION variable to pass login ID
Hi,
I have been trying to solve the problem of using session variables, but I
have not had any luck. What I want to do is simple, I want to set my userid
and password in a login screen and use it later (in another php form) to log
into the database.
In my login PHP file I have the following:
if(isset($_POST['userid']) isset($_POST['pword'])){
$_SESSION['user'] = $_POST['userid'];
$_SESSION['password'] = $_POST['pword'];
}
In my connect to database PHP file I have:
if(isset($_SESSION['user'])){
$user=$_SESSION['user'];
echo Print $user;
}else{
echo Print Missing User;
}
The message I always get is Print Missing User, so I must assume the
global variable is not working. Can anyone help me out here?
Thanks,
Larry
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Re: [PHP] SESSION variable to pass login ID
Hi,
Monday, January 13, 2003, 1:04:49 AM, you wrote:
WG Hi,
WG I have been trying to solve the problem of using session variables, but I
WG have not had any luck. What I want to do is simple, I want to set my userid
WG and password in a login screen and use it later (in another php form) to log
WG into the database.
WG In my login PHP file I have the following:
WG if(isset($_POST['userid']) isset($_POST['pword'])){
WG $_SESSION['user'] = $_POST['userid'];
WG $_SESSION['password'] = $_POST['pword'];
WG }
WG In my connect to database PHP file I have:
WG if(isset($_SESSION['user'])){
WG $user=$_SESSION['user'];
WG echo Print $user;
WG }else{
WG echo Print Missing User;
WG }
WG The message I always get is Print Missing User, so I must assume the
WG global variable is not working. Can anyone help me out here?
WG Thanks,
WG Larry
Make sure you have session_start() at the begining of the second file.
--
regards,
Tom
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RE: [PHP] SESSION variable to pass login ID
Tom,
I do have a session_start() in both files, but it does not seem to help.
- Larry
-Original Message-
From: Tom Rogers [mailto:[EMAIL PROTECTED]]
Sent: Sunday, January 12, 2003 11:36 AM
To: Willie G
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] SESSION variable to pass login ID
Hi,
Monday, January 13, 2003, 1:04:49 AM, you wrote:
WG Hi,
WG I have been trying to solve the problem of using session variables, but
I
WG have not had any luck. What I want to do is simple, I want to set my
userid
WG and password in a login screen and use it later (in another php form) to
log
WG into the database.
WG In my login PHP file I have the following:
WG if(isset($_POST['userid']) isset($_POST['pword'])){
WG $_SESSION['user'] = $_POST['userid'];
WG $_SESSION['password'] = $_POST['pword'];
WG }
WG In my connect to database PHP file I have:
WG if(isset($_SESSION['user'])){
WG $user=$_SESSION['user'];
WG echo Print $user;
WG }else{
WG echo Print Missing User;
WG }
WG The message I always get is Print Missing User, so I must assume the
WG global variable is not working. Can anyone help me out here?
WG Thanks,
WG Larry
Make sure you have session_start() at the begining of the second file.
--
regards,
Tom
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Re: [PHP] SESSION variable to pass login ID
This statement doesnt look right to me.
if(isset($_POST['userid']) isset($_POST['pword'])){
$_SESSION['user'] = $_POST['userid'];
$_SESSION['password'] = $_POST['pword'];
}
the if should be:
if ((statement)(statement))
{
}
so:
if ((isset($_POST['userid'])) (isset($_POST['pword'])))
{
$_SESSION['user'] = $_POST['userid'];
$_SESSION['password'] = $_POST['pword'];
}
That should work.
Willie G [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Tom,
I do have a session_start() in both files, but it does not seem to help.
- Larry
-Original Message-
From: Tom Rogers [mailto:[EMAIL PROTECTED]]
Sent: Sunday, January 12, 2003 11:36 AM
To: Willie G
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] SESSION variable to pass login ID
Hi,
Monday, January 13, 2003, 1:04:49 AM, you wrote:
WG Hi,
WG I have been trying to solve the problem of using session variables,
but
I
WG have not had any luck. What I want to do is simple, I want to set my
userid
WG and password in a login screen and use it later (in another php form)
to
log
WG into the database.
WG In my login PHP file I have the following:
WG if(isset($_POST['userid']) isset($_POST['pword'])){
WG $_SESSION['user'] = $_POST['userid'];
WG $_SESSION['password'] = $_POST['pword'];
WG }
WG In my connect to database PHP file I have:
WG if(isset($_SESSION['user'])){
WG $user=$_SESSION['user'];
WG echo Print $user;
WG }else{
WG echo Print Missing User;
WG }
WG The message I always get is Print Missing User, so I must assume the
WG global variable is not working. Can anyone help me out here?
WG Thanks,
WG Larry
Make sure you have session_start() at the begining of the second file.
--
regards,
Tom
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RE: [PHP] SESSION variable to pass login ID
Hi,
I just found my mistake. I added some debug logic, and found that the
$_POST logic was in the calling form, not the called form. As soon as I
moved the if(isset($_POST['userid']) isset($_POST['pword'])){ to the
second form, everything started to work. Thanks to everyone who responded.
- Larry
-Original Message-
From: Nova [mailto:[EMAIL PROTECTED]]
Sent: Sunday, January 12, 2003 3:04 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] SESSION variable to pass login ID
This statement doesnt look right to me.
if(isset($_POST['userid']) isset($_POST['pword'])){
$_SESSION['user'] = $_POST['userid'];
$_SESSION['password'] = $_POST['pword'];
}
the if should be:
if ((statement)(statement))
{
}
so:
if ((isset($_POST['userid'])) (isset($_POST['pword'])))
{
$_SESSION['user'] = $_POST['userid'];
$_SESSION['password'] = $_POST['pword'];
}
That should work.
Willie G [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Tom,
I do have a session_start() in both files, but it does not seem to help.
- Larry
-Original Message-
From: Tom Rogers [mailto:[EMAIL PROTECTED]]
Sent: Sunday, January 12, 2003 11:36 AM
To: Willie G
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] SESSION variable to pass login ID
Hi,
Monday, January 13, 2003, 1:04:49 AM, you wrote:
WG Hi,
WG I have been trying to solve the problem of using session variables,
but
I
WG have not had any luck. What I want to do is simple, I want to set my
userid
WG and password in a login screen and use it later (in another php form)
to
log
WG into the database.
WG In my login PHP file I have the following:
WG if(isset($_POST['userid']) isset($_POST['pword'])){
WG $_SESSION['user'] = $_POST['userid'];
WG $_SESSION['password'] = $_POST['pword'];
WG }
WG In my connect to database PHP file I have:
WG if(isset($_SESSION['user'])){
WG $user=$_SESSION['user'];
WG echo Print $user;
WG }else{
WG echo Print Missing User;
WG }
WG The message I always get is Print Missing User, so I must assume the
WG global variable is not working. Can anyone help me out here?
WG Thanks,
WG Larry
Make sure you have session_start() at the begining of the second file.
--
regards,
Tom
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RE: [PHP] SESSION variable to pass login ID
Well, start debugging your code then. Do a print_r() of $_POST and
$_SESSION at different points in your files. Maybe you're losing your
session on a certain page or it's getting reset.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
-Original Message-
From: Willie G [mailto:[EMAIL PROTECTED]]
Sent: Sunday, January 12, 2003 2:53 PM
To: Tom Rogers; Willie G
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] SESSION variable to pass login ID
Tom,
I do have a session_start() in both files, but it does not seem to
help.
- Larry
-Original Message-
From: Tom Rogers [mailto:[EMAIL PROTECTED]]
Sent: Sunday, January 12, 2003 11:36 AM
To: Willie G
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] SESSION variable to pass login ID
Hi,
Monday, January 13, 2003, 1:04:49 AM, you wrote:
WG Hi,
WG I have been trying to solve the problem of using session
variables,
but
I
WG have not had any luck. What I want to do is simple, I want to set
my
userid
WG and password in a login screen and use it later (in another php
form)
to
log
WG into the database.
WG In my login PHP file I have the following:
WG if(isset($_POST['userid']) isset($_POST['pword'])){
WG $_SESSION['user'] = $_POST['userid'];
WG $_SESSION['password'] = $_POST['pword'];
WG }
WG In my connect to database PHP file I have:
WG if(isset($_SESSION['user'])){
WG $user=$_SESSION['user'];
WG echo Print $user;
WG }else{
WG echo Print Missing User;
WG }
WG The message I always get is Print Missing User, so I must assume
the
WG global variable is not working. Can anyone help me out here?
WG Thanks,
WG Larry
Make sure you have session_start() at the begining of the second file.
--
regards,
Tom
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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