Babal,
Okay, yeah, I missed the $. So, now everything is working good. Here's what
I did with conditionals depending on type of member and status. Thanks for all
your help.
elseif ($type == "reviewer")
{
$sql = "SELECT status FROM membership
WHERE loginName = '$fusername'";
$result4 = mysql_query($sql) or die ("Couldn't select status.");
$row = mysql_fetch_assoc($result4);
extract($row);
if ($status == "confirmed")
{
$_SESSION['auth']="yes";
$logname=$_POST['fusername'];
$_SESSION['logname'] = $logname;
$today = date("Y-m-d h:i:s");
$sql = "INSERT INTO Login (loginName,loginTime)
VALUES ('$logname','$today')";
mysql_query($sql) or die("Can't execute query.");
header("Location: /reviewer/home.php");
}
else
{
unset($_GET['do']);
$message = "The Login Name, '$_POST[fusername]'
exists, but is still pending confirmation as
a reviewer.<p><br>";
include("login_form.php");
}
}
Again, thanks everyone for your help with this issue. All is good now and my
signin/signup process is complete.
J
jadu_babal <[EMAIL PROTECTED]> wrote:
HI Jeff
here is the error
replace $row = mysql_fetch_assoc($result3); on line 42 with
$row = mysql_fetch_assoc(result3);
here only a $ missing in mysql_fetch_assoc in parameter list.
now try it i think it will work.
bye
Babal
--- In [email protected], Jeffrey Cook <[EMAIL PROTECTED]> wrote:
>
> Babal,
>
> I got this error after typing in your suggestion, perhaps I did
something else wrong:
>
> Warning: mysql_fetch_assoc(): supplied argument is not a valid
MySQL result resource in
/data/11/0/109/110/761599/user/778876/htdocs/Login.php on line 42
>
> Here's my code:
>
> if ($num2 > 0) // password is correct
> {
> $sql = "SELECT loginName FROM membership
> WHERE loginName='$_POST[fusername]'
> AND password=password('$_POST[fpassword]')
> AND type='$type'";
> $result3 = mysql_query($sql) or die ("Couldn't execute
query 3.");
>
> $row = mysql_fetch_assoc(result3); // line 42!
> if ($row['type'] == "standard")
> {
>
> Jeff
>
>
> jadu_babal <[EMAIL PROTECTED]> wrote:
> Hi Jeff
>
> in ur code u are comparing this if ($thetype == "standard")
>
> if i got ur purpose correctly
>
> $thetype this is the variable which u are using for fetching the
> records, now u have to compare it with databse values.
>
> but u r comparing it with the variable $thetype
>
> insteed of this u should use
>
> $row = mysql_fetch_assoc(result3);
> and compare it as
>
> if ($row['type'] == "standard")
>
> try the code i thing it will work and solve ur problem.
>
> bye
> BABAL
>
> --- In [email protected], Jeffrey Cook <jnoelcook@> wrote:
> >
> > Pam,
> >
> > I agree with you. I do not believe we are discussing a design
> issue. I have already finished the design and have built the
> environments and it is working. So I wish to continue discussing my
> coding issues publically in this forum. That's what groups like these
> are for, for sharing code.
> >
> > Okay, so I have got this problem working by creating different
> environments, copies of each page for the different type of membership
> and one for when the member is signed out. All the "signed out" users
> are taken to .htm files and the logged in members are taken to .php
> files. I had a drop down box below the username and password for
> login as a certain type. If they selected STANDARD type and submitted
> the form, they would be taken to the STANDARD environment. But, that
> was only for testing purposes...and I had it working. But I want the
> code to check from the database which type the member signed up as and
> take them automatically. It is not working, but I have found the
> location of the problem...it is in these lines:
> >
> > > $sql = "SELECT loginName FROM membership
> > > WHERE loginName='$_POST[fusername]'
> > > AND type='$thetype' ";
> > > $result3 = mysql_query($sql)or die("Couldn't execute query 3.");
> > >
> > > if ($thetype == "standard")
> >
> > The code pulls the username (loginName) and membership type (type)
> from the membership table exactly as they were working when the user
> selected the type manually. However, for some reason, the code simply
> skips over the last statement above and does not see that $thetype and
> standard are equal and goes to the else, which is not included above
> but follows.
> >
> > I am looking at it and trying many different things, but I still
> can't seem to pull it off.
> >
> > Can anyone help? Did I type a command incorrectly?
> >
> > Thanks for any help.
> >
> > Jeff
> >
> > Pam <ltf01@> wrote:
> > Well, I wouldn't want anyone yelling at you either. Its no problem.
> > I was just thinking though that mingling php with html in a file is
> as much
> > a part of learning php as the code itself. Or at least it is
> important with
> > .asp.
> >
> > Take care,
> > Pam
> >
> > > -----Original Message-----
> > > From: [email protected] [mailto:[EMAIL PROTECTED]
> > > Behalf Of Gerry Danen
> > > Sent: Sunday, June 04, 2006 4:15 PM
> > > To: [email protected]
> > > Subject: Re: [php_mysql] Question about Login / Signout...
> > >
> > >
> > > Pam, I have no problem doing that on
> > > http://groups.yahoo.com/group/php_and_mysql/ because I own that
> list. On
> > > this list it might be considered off-topic, and I don't need
> > > someone yelling
> > > at me. :)
> > >
> > > Gerry
> > >
> > > On 6/4/06, Pam <ltf01@> wrote:
> > > >
> > > > Hi Gerry,
> > > > I'm new to the list, and very new to php. I have learned a lot
> just from
> > > > following everyone's conversations and would really like it if
> you kept
> > > > this
> > > > on list if it is ok with everyone else because I feel like I
> could learn
> > > > from it.
> > > >
> > > >
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> > >
> > > The php_mysql group is dedicated to learn more about the
> > > PHP/MySQL web database possibilities through group learning.
> > > Yahoo! Groups Links
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> >
> >
> >
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