PHP - How Do I Keep $_session Data When Working With Paypal?

The below should choose the radio button that is set by the $_SESSION variable or default to 3. However, it's defaulting to 1.

Code: [Select]
$content.='
<p class="form_item"><label>Product Rating:</label><br />
<div class="rating_radio"><input type="radio" name="review_product_rating" value="1"';
if(isset($review_product_rating) && $review_product_rating=="1"){$content.='checked';}
$content.=' /> <br />1</div>
<div class="rating_radio"><input type="radio" name="review_product_rating" value="2"';
if(isset($review_product_rating) && $review_product_rating=="2"){$content.='checked';}
$content.=' /> <br />2</div>
<div class="rating_radio"><input type="radio" name="review_product_rating" value="3"';
if(isset($review_product_rating) && $review_product_rating=="3" || !isset($review_product_rating)){$content.='checked';}
$content.=' /> <br />3</div>
<div class="rating_radio"><input type="radio" name="review_product_rating" value="4"';
if(isset($review_product_rating) && $review_product_rating=="4"){$content.='checked';}
$content.=' /> <br />4</div>
<div class="rating_radio"><input type="radio" name="review_product_rating" value="5"';
if(isset($review_product_rating) && $review_product_rating=="5"){$content.='checked';}
$content.=' /> <br />5</div>
<div class="worst">(Worst)</div><div class="best">(Best)</div>
</p>

I'm figuring it's because it's multiple if statement (with no elseif) so it's evaluating the first if and being set to true. I tried changing everything else after the first one to elseif, and it's throwing an unexpected T_ELSEIF error.

This problem happens in both IE9 and Chrome 13.0.782.112 m on Windows

If I close the browser down, my session data is gone on browser restart, if however I just close the tab leaving other tabs say with BBC news and restart my app in a new tab, both browsers remember the old session data.

Since the intended user could do this at any time before completing a transaction, I don't actually get a chance to destroy the session this only happens if they complete payment for their order via..
Code: [Select]
  session_destroy();
  $conn->commit();
  $conn->autocommit(TRUE);

In an attempt to clean up I tried various combinations at the top of the start page "index.php"..

Code: [Select]
  session_start();
//  $_SESSION = array();
  session_destroy($_SESSION['cart']);
  session_destroy($_SESSION['address']);
//  $_SESSION['cart'] = array();
//  $_SESSION['address'] = array();
//  session_destroy();

Can anyone please shed any light on this behaviour difference?

Jamie

I am having problems storing a $_POST variable from a form into a $_SESSION.  When I go to the next page and echo out the $_SESSION variable its blank

Page 1 code (page with self submitting form action)


<?php
session_start();

$email = $_POST['email];
$_SESSION['email'] = $email;
?>


Page 2 (page which should be echoing out the value of the session variable)

<?php
session_start();
echo "Your Email Address which you submitted was: " . $_SESSION['email'];
?>


When I echo that out on page 2 its blank

Why is the variable value not carrying over to the 2nd page?

Chad

For some reason I can't get this script working for my IPN.  Please help.


<?php

include("init.inc.php");

$send = 'cmd=_notify-validate&' . http_build_query($_POST);
 
$head  = 'POST /cgi-bin/webscr HTTP/1.0' . "\r\n";
$head .= 'Content-Type: application/x-www-form-urlencoded' . "\r\n";
$head .= 'Content-Length: ' . strlen($send) . "\r\n\r\n";

$fp = fsockopen('sandbox.paypal.com', 80);

if ($fp !== false){
fwrite($fp, $head . $send);
$response = stream_get_contents($fp);
$response = trim(end(explode("\n", $response)));

if ($response === 'VERIFIED' && $_POST['payment_status'] === "Completed" && $_POST['mc_gross'] === "10.00" && $_POST['mc_currency'] === "USD")
{
$payer_email = $_POST['payer_email'];
$q = mysql_query("UPDATE `users` SET `test` = '1' WHERE `email` = '$payer_email'");
}
}
 
fclose($fp);

?>

Hi all,

This is driving me insane!

Basically I am simply trying to setup a reoccurring payment, in live and sandbox I get an error once I submit login details
Code: [Select]
The link you have used to enter the PayPal system is invalid. Please review the link and try again.
The code I am sending to paypal is:
header("location:
https://www.sandbox.paypal.com/cgi-bin/webscr?
cmd=_xclick-subscriptions&
business=sandbox@minecraftclan.com&
item_name=Item1&
a3=3.00&
p3=30&
t3=D");

My sandbox account is 'sandbox@minecraftclan.com', I really cant see what the issue is!

It will show the payment screen, but error occours after logging in with sandbox details, the same if I run to the live site and use an real paypal account

Any ideas?

EDIT:

This iswhat a view source on the error page looks like:
s.prop1="xpt/Merchant/hostedpayments/Error";
s.prop6="6FW885512W222534S";
s.prop7="Personal";
s.prop8="Verified";
s.prop9="Unrestricted";
s.prop10="US";
s.prop14="The link you have used to enter the PayPal system is invalid. Please review the link and try again.";
s.prop16="";
s.prop34="PayPalCredit:Servicing:CO:NoTransactions";
s.prop15="";
s.pageName="xpt/Merchant/hostedpayments/Error::_flow";
s.prop50="en_US";
s.prop18="";

That mean anything to anyone?!

First let me explain my code.
It comprises of three php files.

inc_fn_header_and_menu.php,  contains the HTML and CSS header details and it initializes the session via session_start();

 

This  is later included in project_status.php] .

 

In project_status.php] , I have included another file project_status_app.php which contains a HTML form.

 project_status.php:

<?php
 include 'inc_fn_header_and_menu.php';
function includeFile($file,$variable) {
    $var = $variable;
    include($file);
}
if (isset($_GET['id']) && $_GET['id']!="") {
    $pid    = $_GET['id'];
    $_SESSION['pidForApproval'] = $_GET['id'];
    $query = 'SELECT * FROM `profile` WHERE pid ='.'\''.$pid.'\'';
    $result=mysqli_query($db,$queryToRetrievePP) or die("There are no records to display ... \n" . mysqli_error());
    foreach ($result as $row) {
        $status = $row['status'];
    }        
}
...........some PHP and HTML code.......
	<div id="customerPurchaseApprovalForm">
<?php  
  echo '<p>APPROVAL FOR CUSTOMER PURCHASE</p>';
  $discountApprovalStatus = "Granted";
  if ($discountApprovalStatus == "Granted") {  
 
  includeFile("project_status_app.php",$highestannualvalue);
  } else {
     //.........
  }

In project_status_app.php I am attempting to retrieve pidForApproval from the $_SESSION array.

<?php
        // put your code here  UPDATE `pp` SET `customer_purchase_remarks` = 'hahaha' WHERE `pp`.`id` = 207;
        
        if ($_SERVER['REQUEST_METHOD'] == 'POST') {
            include '../../inc/fastlogin.php';      
            $sql = "UPDATE pp SET customer_purchase_remarks ='{$_POST['remarkstxt']}' WHERE pp.pid='{$_SESSION['pidForApproval']}'"; 
            $result = mysqli_query ( $fastdb, $sql ) ;
            if (mysqli_affected_rows($fastdb) != 1) { 
               $_SESSION['err_cpa_rmks'] = "<p>Error while updating WHERE id='{$_SESSION['pidForApproval']}'</p>";
                //echo "<p>Error while updating WHERE id='{$_POST['pidForApproval']}'</p>".mysqli_error($fastdb);                 
            } else { 
                $_SESSION['suc_cpa_rmks'] = "<p>Records was updated successfully.</p>";
                //echo "Records was updated successfully."; 
            }
            header ("location: project_status.php?id="$_SESSION['pidForApproval']);
            exit();
        }    
        
        ?>

When I load  project_status.php,  project_status_app.php is supposed to display the form. Once the user fills in the form the and the submit button has been pressed, the UPDATE statement is supposed to run and then it is supposed to navigate back to project_status.php?id=FA142. But the update is failing and the when the project_status.php is loaded back, the url looks like this http://localhost/fast/project_status.php?id= . The id is empty. It is supposed to be something like this http://localhost/fast/project_status.php?id=FA142. With the id being populated at the 

header ("location: project_status.php?id=".$_SESSION['pidForApproval']);

I suspected that my $_SESSION['pidForApproval'] is not being populated in project_status.php but I echoed back $_SESSION['pidForApproval'] in that file itself and I can see it is being populated. Hence, I suspect that the $_SESSION['pidForApproval'] is not being passed to project_status_app.php. I have already attempted to include session_start(); clause in project_status_app.php but that gives an error, stating that the session has already started, in inc_fn_header_and_menu.php. Can someone help me as to why the $_SESSION['pidForApproval'] is not being passed on to the project_status_app.php file. Thank you.

Edited August 9, 2020 by Robinson
Missing some information.

I'm trying to set up paypal payment processing on my website.  I have godaddy shared hosting (<--mistake) and I'm trying my hardest to not switch hosts.  The payments process fine, it is the listener that I'm having problems with.  Paypal POSTs data to a script for which you set the url to a "listener" script in your settings.  When a payment is made.... they POST the data to the listener script... you encode the data...append it all together plus another variable and send it back to be verified that it matches... Well I'm not too worried about the script working and all that... the problem is I'm not even RECEIVING the post.  I have a simple foreach loop gather and append the data into a string.  I have it set up right now to insert into MYSQL just so I can SEE if it's working....You can check your IPN history and I can see that payments are made but paypal is sending it over and over it's in a "retrying" state with a 408 error.  My server is not sending a 200ok response. 

After 4 days of bashing my head against the wall I figured out that there's a problem with godaddy shared hosting.. If I use a virtual private server with the SAME script from godaddy it works just fine.  But on the shared hosting it blocks it for some reason...After calling paypal and godaddy countless times with them pointing their finger at each other... one godaddy support rep tells me that I should set my script up with a proxy and he gave me info which I will show you in the script... This is what I need help with right.  How can I receive POST data through a proxy in my script....

I want to add that this doesn't make sense to me because in my script I also have the word "yes" insert into MYSQL even if the script is hit at all... All this time my script is NEVER even touched by paypal... So.. if I add this proxy stuff... how will let my script even be accessed? I'm not sure how servers work together too well to process things.  It just didn't seem like the proxy thing would even help when my script can't even be accessed it seems.  Any help is appreciated. 

Here's the simple code I'm working with:
$con = mysql_connect("","","") or die(mysql_error());
$db = mysql_select_db("swellshirt",$con);

$ip = "64.202.165.130"; // proxy IP <-- IP...
$port = 3128; // proxy port
$url_proxy = 'http://proxy.shr.secureserver.net';



// read the post from PayPal system and add 'cmd'
$req = 'cmd=_notify-validate';

foreach ($_POST as $key => $value) {


$value = urlencode(stripslashes($value));
$req .= "&$key=$value";


}

mysql_query("INSERT INTO orders (description, street) VALUES ('$req', 'yes') ")
or die(mysql_error());

Here is something i dreamt up, i know there is a way for this to work, but i'm not having any great ideas at this point. I want to be able to auto increment through the $_POST[] array data sent from a PayPal IPN. Sample return data looks like

item_number1=val
item_name1=val

so on and so forth. The idea is: While there are $_POST item_numbers, do something with it. The loop continues until there are no more item_numbers. I don't want to use the static methods eg: explicitly defining item_number1, item_number2, ect. This is what i thought would work, but keeps failing -> the script doesn't enter the first condition:

$x = 1; // set the initial item number
        while(isset($_POST['item_number$x'])){
            $qty = $_POST['quantity$x'];
            while($qty > 0){                
                // step through each item response result, setting each item to paid
                $package_id = $_POST['item_number$x'];    
                update_paid_status($transaction_id, $package_id);
                $qty--;            
            }
            $x++;
        }

Any Ideas?

Hello everyone, I am working on a form that is similar to a shopping cart system and I am thinking of creating a button that submits the checked value and saves them to a $_SESSION variable. And also a link that links to a cart.html that takes the values of a $_SESSION variable. I am have trouble figuring what tag/attribute should I use in order to achieve that.   Right now my code attached below submits the checked values to cart.html directly. However I want my submit button to save the checked box to a $_SESSION variable and STAY on the same page. And then I will implement a <a> to link to the cart.php.   I researched a little bit about this subject and I know it's somewhat related to ajax/jquery. I just wanted to know more about it from you guys. I appreciate your attention for reading the post and Thanks!   Below is the form that I currently have:

<form name= "finalForm" method="POST" action="cart.php">
<input type="Submit" name="finalSelected"/>

<?php foreach($FinalName as $key => $item) {?>
<tr>
<td><input type="checkbox" name="fSelected[]" value="<?php echo htmlspecialchars($FinalID[$key])?>" />
<?php echo "$FinalID[$key] & $item";?>
</td>
</tr>
<?php } ;?>

Below is the code for cart.php

<?php
require ('connect_db.php');
if(isset($_POST['finalSelected']))
{
	if(!empty($_POST['fSelected']))
	{
		$chosen = $_POST['fSelected'];
		foreach ($chosen as $item)
			echo "aID selected: $item </br>";
		$delimit = implode(", ", $chosen);
		print_r($delimit);
	}

}
if(isset($delimit))
{
$cartSQL = "SELECT * from article where aID in ($delimit)";
$cartQuery = mysqli_query($dbc, $cartSQL) or die (mysqli_error($dbc));
while($row = mysqli_fetch_array($cartQuery, MYSQLI_BOTH))
{
	$aTitle[] = $row[ 'name' ];
}
}
?>

<table>
	
		<?php if(isset($delimit)) { 
		$c=0; foreach($aTitle as $item) {?>
		<tr>
		<td>
			<?php echo $aTitle[$c]; $c++;?>
		</td>
		</tr>
		<?php }}?>
</table>

Hi,

I am updating a few parts of my current website.
Currently, the website updates user accounts by uploading CSV data reports.

I currently use fgetcsv to deal with each row and this is taking forever to deal with the data as this is currently the code for each row.


while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
{
$grabinfo = "SELECT * FROM transactions WHERE id = '$data[$i]'";
$grabinforesult = mysql_query($grabinfo) or die(mysql_error());
while($row = mysql_fetch_array($grabinforesult)){
$merchant = $row['merchant'];
$username = $row['username'];
$datetime = $row['datetime'];
$letthem = $row['letthem'];
$validate = "SELECT * FROM transactions WHERE merchant = '$merchant' and username = '$username' and (status='pending' or status='success' or status='confirmed') and datetime='$datetime'";
$validator = mysql_query($validate) or die(mysql_error());
if(mysql_num_rows($validator) >= $letthem)
{
$import="UPDATE transactions SET amount='0.00' WHERE id='$data[$i]' and status ='awaiting'";
mysql_query($import) or die(mysql_error());
}}


 

mysql_query($import) or die(mysql_error());
} fclose($handle);


This is performing lots of checks on the csv data and checking it with the system. It checks each row.
Is there a quicker way of reading the csv data? There are lots of checks and while loops within the fgetcsv while loop. It can take upto half an hour to process a file with 1000 csv rows.

Hope someone could offer some advise.

Thanks

Hi,

I need to rack some peoples brains! 

I have here a script that allows a GSM/ Remote receipt printer to poll for data in a specific format. The issue is that I will be deploying a few 100 of these damn things and need to know what ones are on or off! I can get the time I polls the file and have it writing to a DB table. The only issue is that even with only two printers running its filling up the table fast! Over 600 rows in 30 mins for only two printers!  I need to find out a better way to do this but have the same flexibility in knowing what exact times it was on and off and then to take this data and put into a nice viewable format! This should be safe with a few hundred printers at once. 

Hope someone can help me with this as I am totally stumped! =/ 

//database
create table mydata (
	id int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
	fname varchar(20),	
    phoneno int(12) NOT NULL
    /*......*/
);
//class my data php

<?php

include('connect.php');
class InsertMydata {
    public function insertnow($fname, $phoneno) {
        $connect = new Connect;
        $insrt = $db -> prepare('INSERT INTO mydata (fname, phoneno) VALUES (?,?)');
		$insrt -> execute(array($fname, $phoneno));		
	}
}
?>


//insernow validate form

<?php
include('../classs/mydata.php');
//Declare data and error arrays
$errors = [];
$mydara = [];
if(!preg_match('/^[a-zA-Z]{4,15}$/', $_POST['fname'])) {
	$errors['fname'] = 'Enter full name!';
}
//this block not working even the phone exist
$connect = new Connect;
$phoneno = $_POST['phoneno'];
$checkiexist = $connect -> prepare('SELECT * FROM mydata WHERE phoneno = ?');
$checkiexist -> execute([$phoneno]);
if($checkiexist->rowCount() > 0) {
      $errors['phonenoexist'] = 'Try another phone number!';
 }
if(!empty($errors)){
      $data['success'] = false;
      $data['errors'] = $errors;
}else{
      $data['success'] = true;
      $data['message'] = 'success message!';  
      $mydata = new InsertMydata;
      $mydata -> insertnow($fname, $phoneno);
}
echo json_encode($data);

?>

//my ajax

$("#insertbtn").click( function(e) {
     var fname = $('#fname').val(),
         phoneno = $('#phoneno').val(),
           
       $.ajax({
           url: 'insertnow.php',
           type: 'POST',
           data: {fname:fname, phoneno:phoneno},
           dataType: "JSON",
           encode: true,
        }).done( function (data) {
         
         if (data.success == false) {
           if (data.errors.fname) {
                 $('#fname').append('<p class="text-danger">' + data.errors.fname + '</p>');
            } 
            if (data.errors.phonenoexist) {
              $('.card-header').append('<div class="alert alert-info alert-dismissible" role="alert">
                <button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>'+data.errors.phonenoexist+'</div>');
            }
            } else {
             $('.card-header').append('<div class="alert alert-success alert-dismissible" role="alert"><button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>'+data.message+'</div>');
        }
    });
       e.preventDefault();
    });


//the problem is, the code insert data even if the phone exist why?

the problem is, the code insert data even if the phone exist why?

Edited April 8 by mahenda

Hello there,

I have a script which uses the LOAD DATA LOCAL INFILE command see below:

<?php

//connect to your database
mysql_connect("localhost", "xxx", "xxx"); //(host, username, password)

//specify database 
mysql_select_db("xxx") or die("Unable to select database"); //select which database we're using

// Build SQL Query  
$query = "LOAD DATA INFILE  '/public_html/admin/files/test-jp-stock.csv' INTO TABLE 'jpaero_stocksearch' FIELDS TERMINATED BY  ',' LINES TERMINATED BY  '\r\n' IGNORE 1 LINES";
     
if(mysql_query($query)){

echo ">> New Stock Data has now been uploaded. Database is now live and searchable.";}
else{

echo "Upload failed. Please contact support.";}

?> 

Initially I uploaded the csv file through phpmyadmin and everything worked fine - so I then used the SQL generated inside my script, changing the location to where the file actually is - however now nothing happens at all.

Any ideas on whats gone wrong, gratefully received!

Thanks,

Hey guys,

New to the forum and a newer user of PHP / MySQL.  I am having trouble with some code I've written up.  I don't seem to get any errors when running it, but it's not updating my database the way that it should.  hopefully a simple fix.  I am thinking that it must be on the MySQL side of things. 

Couple of things to start. 

My html form is comprised completely of drop down list inputs.  I'm the only user so I thought this would be the easiest approach.

Because of that I've made my PHP as follows:

Code: [Select]
<?php

$season = $_POST['season'];
$month = $_POST['month'];
$day = $_POST['day'];
$year = $_POST['year'];
$time = $_POST['time'];
$event = $_POST['event'];
$game = $_POST['game'];
$buyin = $_POST['buyin'];

$connect = mysql_connect('localhost','root','') or die('can not connect');

if ($connect)
{
    echo "connected to database";
}

$db = mysql_select_db('dpl') or die('can not find database');

if ($db)
{
    echo "DPL Selected";
}

$query = sprintf("INSERT INTO events (season , month , day , year , time , event , game , buyin)
    VALUES ('%s' , '%s' , '%s' , '%s' , '%s' , '%s' , '%s' , '%s')",
       $season ,
       $month ,
       $day ,
       $year ,
       $time ,
       $event ,
       $game ,
       $buyin
       );

if ($query)
{
    echo "Your event has been added";
}

?>
My connection is working, my database is selected and I'm even now getting confirmation that my query is working, but when i go to check my database there are no entries in it?

any thoughts?

I've tried the drop down variables as both VARCHAR and TEXT inputs in MySQL, but I can't seem to get it to work.

Any help is greatly appreciated.