PHP - How To Send Cookies To Another Site

I am working on a phpmailer script that sends an order confirmation email to the customer AND the client at the same time.  If I have the customer email and client email set both to the originating domain's email addresses (myname@myserver.com), then it sends fine.  However, if I try to send to an outside mail server (eg. someone@gmail.com), I get the following errors:

Code: [Select]
SMTP -> FROM SERVER:220 myserver.com ESMTP Exim 4.63 Sat, 18 Sep 2010 15:08:21 -0700
SMTP -> FROM SERVER: 250 myserver.com Hello localhost [127.0.0.1] 250-SIZE 52428800 250-PIPELINING 250-AUTH LOGIN PLAIN 250-STARTTLS 250 HELP
SMTP -> FROM SERVER:250 OK
SMTP -> FROM SERVER:250 Accepted
SMTP -> FROM SERVER:
SMTP -> ERROR: RCPT not accepted from server:
SMTP Error: The following recipients failed: someone@gmail.com Message could not be sent.
Mailer Error: SMTP Error: The following recipients failed: someone@gmail.com

SMTP server error:
I'm not sure what's going on here.  Any SMTP or phpmailer geniuses here that can shed some light on what needs to happen here for this to send to any address?

Hi,

I'm trying to setup a quick PHP script that will grab the email from the url (see below) and after inserting into MySQL db - which is working fine - the script will complete two additional tasks:

1. send that same captured email out to a external db as in shown via http://domain1.com/insert.php?email=$lead (example), but then send to a DIFFERENT source - the originator of the lead - a portback acknowledgement using Header (sending the status and email to http://domain2.com/check.php?e=$lead&s=$status for their records).

See the code below:
-------------------------

Code: [Select]

$lead = $_REQUEST['e_mail'];                // will grab email from posted url string and assign to local variable
$result = mysql_query($command);      // this is just to execute the MySQL insert which works just fine but included here to explain validation below

// Create API Call string to insert lead into iContact folder
$requestURL = "http://domain1.com/insert.php?email=$lead";

// Execute API Call to CAKE
$xml = simplexml_load_file($requestURL) or die("feed not loading");


if ($result) {
    $status = 1;       // mark lead as sucess

    // send postback on lead status
    header("Location: http://domain2.com/check.php?e=$lead&s=$status");
}

--------

Problem: I'm getting all sorts of errors with the simplexml_load_file() function and can't figure out why it won't work.
Any input appreciated as this the only way I know how to pass the lead onward and then inform/update the other party of receipt of information.

thanks!

I have several "sites" located in my html directory, and each has a "general" access point and an "administrator" access point:

/var/www/html/site1/index.php
/var/www/html/site1/administrator/index.php
/var/www/html/site2/index.php
/var/www/html/site2/administrator/index.php
/var/www/html/site3/index.php
/var/www/html/site3/administrator/index.php

All sites are similar except that data will be specific to site1, site2, or site3, etc.   Users who log onto /var/www/html/siteX/index.php are totally unrelated to those who logon to /var/www/html/siteX/administrator/index.php, will have different logon credentials, are stored in different DB tables, and each should have their own session.  If a user logs off of either the general or administrator site, it should not effect the other site even if they were previously logged on to both on the same PC (and of course not effect other sites). When a user logs off, I would like to destroy their previous cookie and associated session.  Users for either will only use https.  I am using Apache to rewrite https://www.mysite.com/ to https://mysite.com/.  While I named the administrator site "administrator" above, the administrator user has the ability to change the directory name.   I am thinking I need to use session_set_cookie_params to specify where I wish the session cookie to be stored since /var/www/html/siteX/administrator/index.php is a sub-directory to /var/www/html/siteX/index.php, but am not really sure.  Sorry for the cryptic post, but I am not very well versed in this subject. How would you recommend setting up cookies/sessions for this scenario?  Thank you

now i use this code to show where the visitors came from to my site.

<?php
$referer=$_SERVER['HTTP_REFERER'];
echo $referer;
?>

now, i want to show the  5 latest vistors referer's site url on my site ?

Not sure if I'm trying to achieve something totally crazy here, or if this is something pretty standard. Didn't have much luck with searching as I'm not fully down with all the terms.

(A) I have one site providing an RSS feed.
(B) I have one site I want to search, once for each of the items in the feed A.
(C) I want the results of the search in (B) to be displayed on page (C).

So for example, the feed on (A) says;

apples
bananas
oranges
cheese

I want site (B) to search for each of those terms (by passing the item in the feed (A) to the ?search= part of the URL of that page) and then show the results from THAT search on page C. Bit of a complex one, let me know if you need me to clarify. Thanks for any help! 

Hi,

My first post here is a cry for help

I have a Windows 2003 server running IIS6/PHP5, the server hosts multiple web sites.

The problem is include files that are for site A are showing on site B (each site having its own includes as part of the site files in its own site folder), though not every time, its very random, sometimes the correct includes show, sometimes ones from another site on the same server. This only occurs where the include files for both sites have the same name, such as 'inc-header.php' for example.

I can only assume PHP is caching includes and because they have the same name is showing the wrong one on other sites sometimes, if I rename them to something unique then the problem goes away, but its not a practical solution to rename all include files to unique names so I find myself looking for a 'real' fix.

I have a feeling its to do with the include_path in the php.ini, but right now its disabled with a semi-colon, and I don't want to set one as I have no global includes, all includes are site specific.

Any help would be very much appreciated!

Phil

I'm currently running a classified ads site and planning to display my own content from database combined with and external site rss. So here is what i got right now after the db query for the jobs ads (procedural php),

while ($row = mysqli_fetch_array($results, MYSQLI_ASSOC)){


echo '<div class="media margin-none">
<a class="pull-left bg-inverse innerAll text-center" href="#"><img src="'.$foto.'" share_alt="" width="100" height="100"></a>
<div class="media-body innerAll">
<h4 class="media-heading innerT">
<a href="' . $row['title'] .'-da' . $row['id_ad'] . '" class="text-inverse">'. $remuneracion .' &nbsp; ' . substr(ucfirst(strtolower($row['title'])), 0, 53) . '</a> <small class="pull-right label label-default"><i class="fa fa-fw fa-calendar-o"></i> ' . $row['date_created'] . '</small></h4>
<p>' . substr(ucfirst(strtolower($row['description'])), 0, 80) . ' ...</p>';

echo '</div>
</div>
<div class="col-separator-h"></div>';
}

echo pagination($statement,$per_page,$page, $url_filtros, $filtros);
?>

it is the while loop that i use to display ads from my database, what could be the best way to display (in this same loop?) other site's rss feed so i can show my content combined with the external rss? Thanks

Transferring data from sub-domain.site.com
Reading sub-domain.site.com

What is this all about?
I'm going to put all .. images into a separate sub-domain eg: images.site.com.
This would create a folder inside my public_HTML called "images"

Now when sites have that Transferring data, and Reading... is this .. something relating to what I want. Facebook also does it, and they get their images for the site from a sub domain, how is it all done?

I'm not sure if its entirely PHP, but I hope someone can help.

Thanks

Hi I made a new design for my website and I made some changes. I want to use layout for my second site. I'll like to know if my site is easier to browse through now and if you like the design better?. I test my site on internet explorer, chrome, and firefox. It is best to use site on better browsers like firefox and chrome to get a better experience of site. Thanks.   http://adjade.com

This is prob a stupid question.. but i've always wondered..   When deleting a cookie why do we use  why do we use: setcookie($name, '', time() - 3600, "/", "", 0);   when this works just fine: setcookie($name, '', 0, "/", "", 0);       isn't the time() just a waste of space?     I ask this because everywhere i look i see: setcookie($name, '', time() -3600 , "/", "", 0);

i was trying to make a new login script and wanted to make point system which i can add points manually to member so i made a new column and named it userpoint i tried many many codes First i used

$_SESSION['userName'] = $username;
echo "Welcome ".$_SESSION['userName']."!";
$_SESSION['userpoint'] = $userpoint;
echo "you got".$_SESSION['userpoint']."!";
 

the page showed username but didn`t show the points ! i tried then i figured out cookies

if(isset($_COOKIE['ID_my_site'])) 

 { 
setcookie( "userpoint");
 	$username = $_COOKIE['ID_my_site']; 
 	$userpoint = $_COOKIE['ID_my_site'];

 	$pass = $_COOKIE['Key_my_site']; 

 	 	$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error()); 
 

 	while($info = mysql_fetch_array( $check)) 	 

 		{  

this time the page shows both of usernames without mentioning the points ! i changed    

$userpoint = $_COOKIE['ID_my_site']; 

 to    

$userpoint = $_COOKIE['Key_my_site'] 

; and it showed the hashed password i want to set a new $_COOKIE so i can get members point to member area is that possible ? the script is still missing secure and salt and the security stuff i`ll add them later i just want to know if is that possible 
Edited by Ahmedamer, 03 September 2014 - 09:39 PM.

The code below is the only part of my script failing and I have no clue why. I even tried setting the TestCookie for 352 days. What boggles my mind is that this code used to work and is now failing. Has something changed?

$value = "wtf";
setcookie("TestCookie",$value, time()+3600*24);

setcookie("user", $name1, time()+14400);
setcookie("userid", $uid1, time()+14400);
setcookie("coid", $coid1, time()+14400);
setcookie("login", "yes", time()+14400);
setcookie("status", $status1, time()+14400);

Okay im having a problem with cookies so if anyone can help i would be grateful.

When you login you can choose how long to stay in for. For testing purposes the choices a

Forever 1 Hour 1 Day Never
Based on your choice i am setting the cookie expiration as follows:

if ($_POST['remember_me'] == '1')
{
setcookie('remember', time() + 99999999999999);
}
elseif ($_POST['remember_me'] == '3600')
{
setcookie('remember', time() + 3600);
}
elseif ($_POST['remember_me'] == '84600')
{
setcookie('remember', time() + 84600);
}
elseif ($_POST['remember_me'] == '0')
{
setcookie('remember');
}
echo $_COOKIE['remember']; 
}

Then for testing I am echoing the cookie at the head of the document:

echo "Cookie: " . @$_COOKIE['remember'];

The problem is that when the browser is closed the cookie is gone. Only the last option "never" is set as a session cookie which means the others
should stay active even when the browser is closed shouldnt they?

Anything i have missed here?

When a user logs in it sets a cookie with their user id and sets the time they choose(either a session cookie or a cookie lasting one year for users who wish to stay logged in). when they select to stay logged in forever and close the browser the next time they open it, it tells them they arent logged in. however when they go to a new page they appear to be logged in. What i dont understand is why they have to go to a new page for it to say they are logged in. Here is the code which runs everytime the site is load


if (isset($_COOKIE['uid']))
{
    $user->setup($_COOKIE['uid']);
}


user setup basically selects their info from the database and sets their username and other info to variables.

Anyone know a better way to do this?