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PHP - Paypal Ipn Return Item Data Dynamically Handled - Item_number1, 2, 3 Ect [code]
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? Similar TutorialsHello, I'm learning how to write simple CMS and I'd like it to support page aliases. So that instead of http://www.myweb.com/?id=123 I could use http://www.myweb.com/some-page for example. I don't want to use rewrite rules in .htaccess but I'd like to do it similarly as Drupal. I understood Drupal redirects 404 page not found error to index.php and then somehow handles it. Can you advise how to do it please? Or is there any tutorial or example available? Thank you in advance! Vojta Hi, I have setup my paypal payment method for a shopping cart on my site, all is well but in order for my server to flag an invoice as paid, the user must return to my website after paying and the query string in the url give the script the go ahead to mark it as paid which in turn alerts the appropriate department that its ok to source the order. ~The problem is, if the user for whatever reason skip the redirect back to my site, the invoice is remain 'unpaid' and has to be changed manually when the payment has been checked in the account. As you might imagine this is a bit of a problem because not only can it cause quite a delay, it also requires manually changing values which can lead to other problems. So if there is a better way i would appreciate the info, thanks a lot. I have somewhat of a dilemma. When someone clicks on a Buy Now button and subsequently follows necessary steps to complete process of purchase, when that transaction is completed, in my product table, I want to subtract 1 from whatever value is in the field. E.g. say product one is being purchased, prior to purchase in product table, there are 5 product one's in stock, when purchase is complete, subtract 1 from 5 (to get 4). Here is my code Code: [Select] <?php $title = "Like This Product, Buy It NOW!!!"; require ('includes/config.inc.php'); include ('./includes/header.html'); require (MYSQL); include ('./includes/main.html'); if($id = isset($_GET['prodID'])) { $query = "SELECT `prodID`, `product`, `prod_descr`, `image`, `price` FROM product WHERE `prodID`='{$_GET['prodID']}'"; $r = mysqli_query($dbc, $query); $showHeader = true; echo "<div id='right'>"; while($row = mysqli_fetch_array($r)) { if($showHeader) { //Display category header echo "<h1>" . "<span>" . "# " . "</span>" . $row['product'] . "<span>" . " #" . "</span>" . "</h1>"; echo "<div id='item'>"; // div class 'item' echo "<div class='item_left'>"; echo "<p id='p_desc'>"; echo $row['prod_descr']; echo "</p>"; echo "<p>" . "<span>" . "£" . $row['price'] . "</span>" . "</p>"; echo "</div>"; echo "<div class='item_right'>"; echo "<img src='db/images/".$row['image']."' />"; $showHeader = false; echo "</div>"; ?> <p> <form target="paypal" action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_s-xclick"> <input type="hidden" name="hosted_button_id" value="7UCL9YCYYXL3J"> <input type="hidden" name="item_name" value="<?php echo $row['product']; ?>"> <input type="hidden" name="item_number" value="<?php echo $row['prodID']; ?>"> <input type="hidden" name="amount" value="<?php echo $row['price']; ?>"> <input type="hidden" name="currency_code" value="GBP"> <input type="image" src="https://www.sandbox.paypal.com/en_US/i/btn/btn_cart_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.sandbox.paypal.com/en_US/i/scr/pixel.gif" width="1" height="1"> </form> </p> <p> <form name="_xclick" action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="business" value="me@mybusiness.com"> <input type="hidden" name="currency_code" value="GBP"> <input type="hidden" name="item_name" value="<?php echo $row['product']; ?>"> <input type="hidden" name="amount" value="<?php echo $row['price']; ?>"> <input type="image" src="http://www.paypal.com/en_US/i/btn/btn_buynow_LG.gif" border="0" name="submit" alt="Make payments with PayPal - it's fast, free and secure!"> </form> </p> <?php echo "</div>"; // End of div class 'item' $strSQL = "SELECT prodID, product, price, image FROM product ORDER BY RAND() LIMIT 1"; $objQuery = mysqli_query($dbc, $strSQL) or die ("Error Query [".$strSQL."]"); while($objResult = mysqli_fetch_array($objQuery)) { echo "<div class='love'>"; echo "<h6>Like this......you'll love this!!!</h6>"; echo "<ul>"; echo "<li>" . "<img src='db/images/" . $objResult['image'] . "' width='50' height='50' />" . "</li>"; echo "<br />"; echo "<li>" . "<a href='item.php?prodID={$objResult['prodID']}' title='{$objResult['product']}'>" . $objResult['product'] . "</a>" . " - " . "£" . $objResult['price'] . "</li>"; echo "</ul>"; echo "</div>"; } } } ?> <?php echo "</div>"; } include ('./includes/footer.html'); ?> How is this achievable please? Hi, For about a month, I have been trying to figure out why my code will not return anything after posting a wwwForm (I have also tried the newer equivalent of this function but I had no luck with that either.) The nameField and passwordField are taken from text boxes within the game and the code used in my login script is copied and pasted from a Register script but I have changed the file location to the login.php file. The register script works fine and I can add new users to my database but the login script only outputs "Form Sent." and not the "present" that should return when the form is returned and it never gets any further than that point meaning that it lets the user through with no consequence if they use an invalid name because the script never returns an answer. What should I do to fix this? Thanks, Unity Code:
using System.Collections;
using UnityEngine;
using UnityEngine.UI;
using UnityEngine.Networking;
public class Login : MonoBehaviour
{
public InputField nameField;
public InputField passwordField;
public Button acceptSubmissionButton;
public void CallLogInCoroutine()
{
StartCoroutine(LogIn());
}
IEnumerator LogIn()
{
WWWForm form = new WWWForm();
form.AddField("username", nameField.text);
form.AddField("password", passwordField.text);
WWW www = new WWW("http://localhost/sqlconnect/login.php", form);
Debug.Log("Form Sent.");
yield return www;
Debug.Log("Present"); if (www.text[0] == '0')
{
Debug.Log("Present2");
DatabaseManager.username = nameField.text;
DatabaseManager.score = int.Parse(www.text.Split('\t')[1]);
Debug.Log("Log In Success.");
}
else
{
Debug.Log("User Login Failed. Error #" + www.text);
}
}
public void Validation()
{
acceptSubmissionButton.interactable = nameField.text.Length >= 7 && passwordField.text.Length >= 8;
}
}
login.php:
<?php
echo "Test String2";
$con = mysqli_connect('localhost', 'root', 'root', 'computer science coursework');
// check for successful connection.
if (mysqli_connect_errno())
{
echo "1: Connection failed"; // Error code #1 - connection failed.
exit();
}
$username = mysqli_escape_string($con, $_POST["username"]);
$usernameClean = filter_var($username, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW | FILTER_FLAG_STRIP_HIGH);
$password = $_POST["password"];
if($username != $usernameClean)
{
echo "7: Illegal Username, Potential SQL Injection Query. Access Denied.";
exit();
}
// check for if the name already exists.
$namecheckquery = "SELECT username, salt, hash, score FROM players WHERE username='" . $usernameClean . "';";
$namecheck = mysqli_query($con, $namecheckquery) or die("2: Name check query failed"); // Error code # 2 - name check query failed.
if (mysqli_num_rows($namecheck) != 1)
{
echo "5: No User With Your Log In Details Were Found Or More Than One User With Your Log In Details Were Found"; // Error code #5 - other than 1 user found with login details
exit();
}
// get login info from query
$existinginfo = mysqli_fetch_assoc($namecheck);
$salt = $existinginfo["salt"];
$hash = $existinginfo["hash"];
$loginhash = crypt($password, $salt);
if ($hash != $loginhash)
{
echo "6: Incorrect Password"; // error code #6 - password does not hash to match table
exit;
}
echo "Test String2";
echo"0\t" . $existinginfo["score"];
?>
This could be PHP or MySql so putting it in PHP forum for now... I have code below (last code listed) which processes a dynamically created Form which could have anywhere from 0 to 6 fields. So I clean all fields whether they were posted or not and then I update the mySQL table. The problem with this code below is that if, say, $cextra was not posted (i.e. it wasnt on the dynamically created form), then this code would enter a blank into the table for $cextra (i.e. if there was already a value in the table for $cextra, it gets overwritten, which is bad). What is the best way to handle this? I'm thinking i have to break my SQL query into a bunch of if/else statements like this... Code: [Select] $sql = "UPDATE cluesanswers SET "; if (isset($_POST['ctext'])){ echo "ctext='$ctext',"; } else { //do nothing } and so on 5 more times.... That seems horribly hackish/inefficient. Is there a better way? Code: [Select] if (isset($_POST['hidden']) && $_POST['hidden'] == "edit") { $cimage=trim(mysql_prep($_POST['cimage'])); $ctext=trim(mysql_prep($_POST['ctext'])); $cextra=trim(mysql_prep($_POST['cextra'])); $atext=trim(mysql_prep($_POST['atext'])); $aextra=trim(mysql_prep($_POST['aextra'])); $aimage=trim(mysql_prep($_POST['aimage'])); //update the answer edits $sql = "UPDATE cluesanswers SET ctext='$ctext', cextra='$cextra', cimage='$cimage', atext='$atext', aextra='$aextra', aimage='$aimage'"; $result = mysql_query($sql, $connection); if (!$result) { die("Database query failed: " . mysql_error()); } else { } Here's the code that deals with the client side:
<?php
session_start();
if(!isset($_SESSION['Logged_in'])){
header("Location: /page.php?page=login");
}
?>
<!DOCTYPE Html>
<html>
<head>
<!--Connections made and head included-->
<?php require_once("../INC/head.php"); ?>
<?php require_once("../Scripts/DB/connect.php"); ?>
<!--Asynchronously Return User Names-->
<script>
$(document).ready(function(){
function search(){
var textboxvalue = $('input[name=search]').val();
$.ajax(
{
type: "GET",
url: 'search.php',
data: {Search: textboxvalue},
success: function(result)
{
$("#results").html(result);
}
});
};
</script>
</head>
<body>
<div id="header-wrapper">
<?php include_once("../INC/nav2.php"); ?>
</div>
<div id="content">
<h1 style="color: red; text-align: center;">Member Directory</h1>
<form onsubmit="search()">
<label for="search">Search for User:</label>
<input type="text" size="70px" id="search" name="search">
</form>
<a href="index.php?do=">Show All Users</a>|<a href="index.php?do=ONLINE">Show All Online Users</a>
<div id="results">
<!--Results will be returned HERE!-->
</div>
search.php
<?php //testing if data is sent ok echo "<h1>Hello</h1><br>" . $_GET['search']; ?>This is the link I get after sending foo. http://www.family-li...php?&search=foo Is that mean it was sent, but I'm not processing it correctly? I'm new to the whole AJAX thing. hey gang: i have a number of variables stored in the $_SESSION array and wanted to know how i can keep them handy for use AFTER PayPal has done it's thing. to help clarify, the desired process here is client fills out form. clicks PayPal button. form data is then used to create some documents which are then sent to client. i have the flow working but without the PayPal part because i'm not sure how to preserve the form data while PayPal does it's thing so i can use it when PayPal is done. am i making sense? i'd like to steer away from client-side cookies because of security issues and our paranoid client insists on it... bah. first the JSON shuffle from JavaScript to PHP and now this PayPal mess. any help is MOST welcome. TIA! WR! 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?! 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()); Hi guys, im trying to connect to a database and get the value for the user in the row called 'user_credit', if it equals 1 or more then i want to show the ''You have £ ....'' bit in the script. Problem is nothing shows at all, even without the if statement. I have changed the value for me in the database so in user_credit the value is 100, which is more than 1 so it should appear. I have probably done something wrong. Any ideas? Code: [Select] <? include '../admin/database/membership_dbc.php'; $r = mysql_query("SELECT * FROM users WHERE user_name='".safe($_SESSION['user_name'])."'") or die ("Cannot find table"); while( $cred = mysql_fetch_array($r) ) { if ($cred >= '1' ) { ?> <p>You have £<? echo $cred['user_credit']; ?> available on you account, would you like to use it on this order?<br> <label for="credit"></label> <select name="credit" id="credit"> <option value="Y" selected>Yes, use credit</option> <option value="N">No, save credit</option> </select> </p> <? } } ?> This is something I've been trying to figure out for some time. I've read blogs and other forums and am still not clear.
Seems that when I pass a variable that has Apostrophe's in the variable, from a form page to the submit page and insert it into the MySql DB table, it inserts OK without any / before the apostrophe.
On the other hand on the same submit page, there is a select query from another table and there are variables with apostrophe's. These queried variables keep the variables from the form page and the queried DB from inserting into a new table.
So I use mysql_real_escape_string () for the variables queried from the table to be inserted into the new table, don't use mysql_real_escape_string () on the variables passed frm the form page, and everything inserts into the new table just fine. Displays with no forward slashes.
My confusion comes from when to use mysql_real_escape_string (), stripslashes () and htmlspecialchars().
Also in the reading I was doing, it looks like mysql_real_escape_string () is being replaced with mysqli_real_escape_string (), but when I tried to use it on a variable queried from the DB something like
$username = mysqli_real_escape_string ( $s['username'] )( $s being 'foreach ( $result as $s )' from the select query. Thanks in advance for shedding any light on this. Hey all, Let's say if all of the data in the site is the result of queries from the database that are output as html by php as users interact with the site. For example, there's no home.php file. Rather user clicks the link to view the page, and a php script pulls data from database and renders the page at that time. There is no javascript involved in this process. Would this page be crawlable by google? Thanks for response. I have a problem with my rest service as below: i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
Hello, Curious to know if someone could point me in the right direction, been struggling with this for a bit now. I have a HTML page with a search field, I can enter a search term and hit the submit button and I am directed to my search.php page with the appropriate results. What I am looking to accomplish is having the search results from the search.php page displayed in a text area below my search field in my HTML page. I have included an image to better describe what I am looking to accomplish: Additionally below is the source from my HTML page and search.php page: page.html <form name="search" action='search.php' method="post"> <input type="text" class="myinputstyle" name="search" value="search" onClick="this.value=''"/><br> <input type="submit" value="submit" class="myinputstyle"> </form> search.php <?php $search = "%" . $_POST["search"] . "%"; mysql_connect ("localhost", "game_over", "Ge7Ooc9uPiedee3oos9xoh4th"); mysql_select_db ("game_over"); $query = "SELECT * FROM game_over WHERE first_name LIKE '$search'"; $result = mysql_query ($query); if ($result) { while ($row = mysql_fetch_array ($result)) { echo "Name: {$row['name']} " . "{$row['lname']} <br>" . "Email: {$row['email]} <br>" . } } ?> Any insight would be most appreciated. Thank you. I have a script that seems to work well to insert a bookmark into a users database when he/she is logged into the system but I am having a hard time figuring out how I would go about making a work-a-round for having an item selected before being logged in, and inserted after they have logged in or registered. For example, I would like a user to be able to select an Item to add to bookmark whether that user is logged in/registered or not and if they are not, they would be greeted with a login/registration form and after successful login the add bookmark script would be initiated on the item previously selected. What I've got this far: Simple form to add bookmark: <form name="bm_table" action="add_bms.php" method="post"> <input type="text" name="new_url" value="http://" /> <input type="submit" value="Add Bookmark"/> </form> Then I have the add bookmark script: BEGIN php $new_url = $_POST['new_url']; try { check_valid_user(); //cannot get past this part since it ends the script....code below if (!filled_out($_POST)) { throw new Exception('Form not completely filled out.'); } // check URL format if (strstr($new_url, 'http://') === false) { $new_url = 'http://'.$new_url; } // check URL is valid if (!(@fopen($new_url, 'r'))) { throw new Exception('Not a valid URL.'); } // try to add bm add_bm($new_url); echo 'Bookmark added.'; // get the bookmarks this user has saved if ($url_array = get_user_urls($_SESSION['valid_user'])) { display_user_urls($url_array); } } catch (Exception $e) { echo $e->getMessage(); } END php Checking valid user - the portion I cannot get past in the above script: function check_valid_user() { // see if somebody is logged in and notify them if not if (isset($_SESSION['valid_user'])) { echo "Logged in as ".$_SESSION['valid_user'].".<br />"; } else { // they are not logged in do_html_heading('Problem:'); echo 'You are not logged in.<br />'; do_html_url('login.php', 'Login'); do_html_footer(); exit; } } How would I go about modifying the script so that a user could fill in the form (later it would be a link...obviously they probably wouldn't be filling in a form that is log-in specific - but same concept I think) Thanks in advance for the help! tec4 I'm working on this database that was built by somebody else, and I'm having issues with. I'm attaching a photo of the database. The problem I'm having is that even when I hard-code variables in the statement, it still brings back "Sorry, but we can not find an entry to match your query" Is there something that I'm not seeing, or something that I'm doing wrong here? You can clearly see in the photo that I'm trying to grab the info with the id_pk 300000. Code: [Select] <?php //$search = $_POST['search']; //$search = '300000'; $data = mysql_query("SELECT * FROM page_listings WHERE id_pk = '300000'"); while($result = mysql_fetch_array( $data )) { $name = $result['page_name']; $id = $result['id_pk']; $page = $result['html']; $cat = $result['category']; $dir = $result['directory']; echo "ID: " .$id ."<br>Page: ". $name ."<br>HTML: ". $page ."<br>Category: ". $cat ."<br>Directory: ". $dir ."<br />"; } $anymatches=mysql_num_rows($data); if ($anymatches == 0) { echo "Sorry, but we can not find an entry to match your query<br><br>"; } ?> Thanks in advance [attachment deleted by admin] Hi there, I think this is a big question but I'd appretiate any help you can provide!! I have a list of items and subitems in a table that looks like this: id parent_id title 1 0 House Chores 2 1 Take Out Trash 3 1 Clean Room 4 0 Grocery List 5 4 Eggs 6 4 Produce 7 6 Lettuce 8 6 Tomato 9 4 Milk I want to display it like this: (+) House Chores: > Take Out Trash > Clean Room (+) Grocery List: > Eggs (+) Produce > Letutce > Tomato > Milk So basically each entry in the table has an unique id and also a parent id if it's nested inside another item. I "sort of" got it figured out in one way, but it doesnt really allow for nested subgroups. I'd like to know how would y'all PHP freaks to this Also taking suggestions for the javascript code to expand/collapse the tree !! Thank you! Hi, I am facing issue in URL validation. I am using following code which return http code and then I use that code to classify about the valid or invalid URL. I can in database that many urls return code zero while url is correct. For example: http://pickadoll.org >> this url return code zero while this url is correct and accessible http://mammasofia.blogg.se >> this url return code zero while this url is correct and accessible Here is code: if ($url == null) $valid['result'] = false; $ch = curl_init($url); $useragent = "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.1) Gecko/20061204 Firefox/2.0.0.1"; curl_setopt($ch, CURLOPT_USERAGENT, $useragent); curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt ($ch,CURLOPT_VERBOSE,false); curl_setopt($ch, CURLOPT_TIMEOUT, 30); curl_setopt($ch,CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt($ch,CURLOPT_SSLVERSION,3); curl_setopt($ch,CURLOPT_SSL_VERIFYHOST, FALSE); $data = curl_exec($ch); $httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE); curl_close($ch); $valid['httpcode'] = $httpcode; if ($httpcode >= 200 && $httpcode < 400) { $valid[result] = true; } else { $valid['result'] = false; } return $valid; I am getting thousands of urls with code 0 but 0 is not a code where I can put condition that 0 code urls are correct. Any help in this issue is appreciated. Thanks alot. Regard, Maesam Well I am looking to change this url Code: [Select] http://website.com/product.php?Item=2369 to Code: [Select] http://website.com/product.php?Item=Item-Name Heres a snip of the code that handles that. <?php include_once('mysql_connect.php');$id = (int)$_GET['Item'];?>() any help would be appreciated. |
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