[PHP-DB] arrays and email

2002-01-29 Thread Kevin Ruiz

I'm working on an application that will allow someone to view all attendees
for a specific webinar that my company is hosting.  I want to allow the user
to send one group email to all participants scheduled for that particular
webinar.

After I connect to my database my code looks like this:

?
  $sql = select * from webusers where webdate=\$webdate\;
  $result = mysql_query($sql) or die(couldn't generate a list of the
users);

while ($row = mysql_fetch_row($result))
 {
  $real_name = $row[1];
  $email = $row[12];
  $list[] = $email;
 }

   echo form method=\post\ action=\doemailattendees.php\\n;
   echo table width=\100%\ border=0 cellspacing=0 cellpadding=0
class=\orange4\\n;
   echo tr\n;
   echo td valign=\top\pbTo:/b/p/td\n;
   echo td valign=\top\p\n;
   foreach ($list as $value)
 {
 print $value, ;
 $to = $value;
 }
   echo /p/td\n;
  echo /tr\n;

echo /table\n;
echo table width=\100%\ border=0 cellspacing=0 cellpadding=0\n;
  echo tr\n;
   echo td valign=\top\pbSubject:/b/p/td\n;
   echo td valign=\top\pinput type=\text\
name=\subject\/p/td\n;
  echo /tr\n;
  echo tr\n;
   echo td valign=\top\pbMessage:/b/p/td\n;
   echo td valign=\top\textarea
name=\message\/textarea/td\n;
  echo /tr\n;
  echo tr\n;
   echo td colspan=2 valign=\top\input type=\submit\
value=\submit\/td\n;
  echo /tr\n;
echo /table\n;
  echo /form\n;
  ?

The $to, $subject,  $message variables then get sent to a page that
actually mails the message.  The problem I'm having is that it's only being
sent to the last person in the array.  I understand why this is happening
but don't know enough about arrays to find a solution.  As my code shows I
ambitiously tried setting $to to the entire array but that doesn't work.

If anyone would be kind enough to help me out I'd greatly appreciate it.

Thank you.
Kevin

www.worktiviti.com



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[PHP-DB] rewriting the $server_name variable with php?

2002-01-07 Thread Kevin Ruiz

Is there anyway to rewrite the $server_name variable with php?  This is kind
of an ambitious post for I think I already know the answer.

My company has just changed our corporate identity and we want our old url
to point to the new one, http://www.worktiviti.com.  This isn't a problem,
we've changed the DNS entries and everything is fine except for the fact
that my boss doesn't want people to see the old url.  He wants the url to
reflect the new company name, worktiviti.  I've talked to our hosting
company and they said they would look into editing the httpd.conf file but
it doesn't seem likely that they would change the config file for one
customer.  I know that you can change the url through the mod_rewrite module
with Apache but I am not that familiar with apache nor do I have access to
the config file.  If anyway has any suggestions as to how php could help me
out I'd appreciate it.

Thanks in advance,
Kevin



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[PHP-DB] overwriting new information in a table

2001-12-02 Thread Kevin Ruiz

I'm working on an application that will allow resellers to log into the site
and see their potential buyers.

When they log in they see their potential buyers and by clicking a link they
will get more detailed information about their contacts.  They will also
have a link to update their clients listings.

As of now I have this set up like this...

A person logs in and sees their contacts.  If they click the update link
they are takent to a page that lists their contact information.  I've put
the respected variables in the value attributes of the input tags.  I want
to make this process as easy as possible.  I dont' want them to have to
update every field if they just want to change a contact's phone number.

Here's the problem...when they click the button to send the changes they
still see their old information.  I know that this is because I've set the
value to their old variables.  Does anyone know a way to get their old input
in as the variable value and to still be able to see their new information
on the next page.

I also want this information to overwrite any existing record based on the
potential contact's id.  The sql section of the code I have looks like this:
$sql = INSERT INTO $table_name
(id, contactid, fname, lname, title, cnl, address1,
address2, city, state, postcode,
country, phone, cell, fax, emailnl, smanager,
percomplete, nfud, exrevenue, wl, cwl)
VALUES
(\\, \$contactid\, \$fname\, \$lname\,
\$title\, \$cnl\, \$address1\, \$address2\, \$city\, \$state\,
\$postcode\, \$country\, \$phone\,
\$cell\, \$fax\, \$emailnl\, \$smanager\,
\$percomplete\, \$nfud\, \$exrevenue\, \$wl\, \$cwl\)
;

$result = @mysql_query($sql,$connection)
 or die(Couldn't execute query.);

I know this doesn't overwrite each entry.  I've also tried something like
this:
$sql = mysql_query(update my_contacts set fname=$fname,
lname=$lname where id=$id);

Whenever I would try to run that query I would get an error back saying I
couldn't execute that query.

If anyone has any ideas I would be more than grateful.  Thanks for your
time.

I'm new to the php/mysql world so if I didn't provide enough relevant code
let me know.

Thanks.
Kevin



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[PHP-DB] Using variables within a SELECT statement

2001-11-28 Thread Kevin Ruiz

I'm working on an application that selects a user's userid if their password
and login match.  If the login and password match I want my variable $cir to
run its own select statement and return its corresponding contactid...for
this example we'll say that value = 5.  I then want to plug in $cir into my
select statement that will show the user the list of his/her contacts and
only theirs.  This seems to be the select statement that's giving my
problems.

$query = select * from my_contacts where
contactid=$cir;

Whenever I run my script I get an error saying that I have a problem with my
sql syntax.  I know that the problem is with the variable becuase it runs
fine with a value in it.

I've tried writing the $cir as '$cir' and '\$cir' but nothing seems to work.

Can anyone help.

Thanks in advance
Kevin



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[PHP-DB] resource id #2, #3, #4......

2001-11-28 Thread Kevin Ruiz

I've come across yet another problem.

I have a table set up that houses four things, a person's real name,
username, pass, and id.  The id is used to join another table which houses
that persons contacts.  I've been validating the user and pass by comparing
the number of rows that the sql statement returns.

Once the user submits their user/pass they're sent to the validation page
where the user and pass are compared and if they're valid another sql
statement takes place:
 $ci = select contactid from users where username='$username' and
password='$password';
 $cir = mysql_query($ci)or die(Couldn't execute);
 echo(htmlheadMETA HTTP-EQUIV=\refresh\
CONTENT=\0;url=db.php\/head/html);

I've registered the variables $username, $password,  $cir and if the user
is validated they're sent to the contacts page.

On the contacts page I'm trying to select which contacts to show based on
the contact id.  This is what the code looks like on the page where the
contacts are going to be displayed...

mysql_connect(localhost) or
 die(couldn't connect);
mysql_select_db(act) or
 die(couldn't connect to the database);

$ci = select contactid from users where username='$username' and
password='$password';
$cir = mysql_query($ci)
or die(Couldn't execute);
$query = select * from my_contacts where contactid='$cir';

$result = mysql_query($query) or
 die( mysql_error() );

When I try to print $cir to see if anything's getting passed I keep getting
something that reads resource id #2.

If I were rename the variables in the $ci, $cir, and $query lines, run the
code, and try to print $cir again I'd get resource id #3 and so on.

Does anyone know what this means and how I can work around it.

Thanks for your time.
Kevin



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[PHP-DB] Newbie help with cookies/sessions

2001-11-27 Thread Kevin Ruiz

I'm new to the PHP and Mysql world and I hope I have an problem that can be
answered quickly.  It seems simple but who knows.

Whenever I try to use cookies or start a session I get the following error:

Warning: open(/Temp\sess_ba759cb3d78fadb70745ca480f1d8661, O_RDWR)
failed: m (2) in Unknown on line 0
Warning: Failed to write session data (files). Please verify that
the current setting of session.save_path is correct (/Temp) in Unknown on
line 0

I tried checking the session.save_path setting the php.ini but I'm not sure
if this has to saved to a specific directory or just a valid directory.

Any help would be greatly appreciated.

Thank you.
Kevin



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